304 North Cardinal St.
Dorchester Center, MA 02124

# Finally, Non Ad-Hoc Problem CodeChef Solution – Queslers

## Problem: Non Ad-Hoc Problem Code Chef Solution

You are given two integers AA and NN. Calculate number of permutations pp of length NN such that i−N+A<pi<i+Ai−N+A<pi<i+A for 1≤i≤N1≤i≤N. Since this number can be very large, calculate it modulo 998244353998244353.

### Input Format

The only line of the input contains two integers NN, AA.

### Output Format

Output the number of permutations modulo 998244353998244353.

### Constraints

• 2≤N≤1092≤N≤109
• 1≤A≤min(500000,N−1)1≤A≤min(500000,N−1)

### Sample Input 1

4 2


### Sample Output 1

5


### Explanation

We need 1≤p1≤21≤p1≤2, 1≤p2≤31≤p2≤3, 2≤p3≤42≤p3≤4, 3≤p4≤43≤p4≤4.

There are 55 such permutations: (1,2,3,4),(1,2,4,3),(1,3,2,4),(2,1,3,4),(2,1,4,3)(1,2,3,4),(1,2,4,3),(1,3,2,4),(2,1,3,4),(2,1,4,3).

### Sample Input 2

322 228


### Sample Output 2

842567743

## Finally, Non Ad-Hoc Problem CodeChef Solution in C++

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <complex>
#include <deque>
#include <forward_list>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template <typename T, typename V>
void ndarray(vector<T>& vec, const V& val, int len) { vec.assign(len, val); }
template <typename T, typename V, typename... Args> void ndarray(vector<T>& vec, const V& val, int len, Args... args) { vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); }); }
template <typename T> bool chmax(T &m, const T q) { return m < q ? (m = q, true) : false; }
template <typename T> bool chmin(T &m, const T q) { return m > q ? (m = q, true) : false; }
int floor_lg(long long x) { return x <= 0 ? -1 : 63 - __builtin_clzll(x); }
template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template <typename T> vector<T> sort_unique(vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; }
template <typename T> int arglb(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::lower_bound(v.begin(), v.end(), x)); }
template <typename T> int argub(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::upper_bound(v.begin(), v.end(), x)); }
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <typename T, size_t sz> ostream &operator<<(ostream &os, const array<T, sz> &arr) { os << '['; for (auto v : arr) os << v << ','; os << ']'; return os; }
#if __cplusplus >= 201703L
template <typename... T> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; }
template <typename... T> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { os << '('; std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os << ')'; }
#endif
template <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T, typename TH> ostream &operator<<(ostream &os, const unordered_set<T, TH> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << '(' << pa.first << ',' << pa.second << ')'; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
template <typename TK, typename TV, typename TH> ostream &operator<<(ostream &os, const unordered_map<TK, TV, TH> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
#ifdef HITONANODE_LOCAL
const string COLOR_RESET = "\033[0m", BRIGHT_GREEN = "\033[1;32m", BRIGHT_RED = "\033[1;31m", BRIGHT_CYAN = "\033[1;36m", NORMAL_CROSSED = "\033[0;9;37m", RED_BACKGROUND = "\033[1;41m", NORMAL_FAINT = "\033[0;2m";
#define dbg(x) cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << endl
#define dbgif(cond, x) ((cond) ? cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << endl : cerr)
#else
#define dbg(x) (x)
#define dbgif(cond, x) 0
#endif

template <int md> struct ModInt {
#if __cplusplus >= 201402L
#define MDCONST constexpr
#else
#define MDCONST
#endif
using lint = long long;
MDCONST static int mod() { return md; }
static int get_primitive_root() {
static int primitive_root = 0;
if (!primitive_root) {
primitive_root = [&]() {
std::set<int> fac;
int v = md - 1;
for (lint i = 2; i * i <= v; i++)
while (v % i == 0) fac.insert(i), v /= i;
if (v > 1) fac.insert(v);
for (int g = 1; g < md; g++) {
bool ok = true;
for (auto i : fac)
if (ModInt(g).pow((md - 1) / i) == 1) {
ok = false;
break;
}
if (ok) return g;
}
return -1;
}();
}
return primitive_root;
}
int val;
MDCONST ModInt() : val(0) {}
MDCONST ModInt &_setval(lint v) { return val = (v >= md ? v - md : v), *this; }
MDCONST ModInt(lint v) { _setval(v % md + md); }
MDCONST explicit operator bool() const { return val != 0; }
MDCONST ModInt operator+(const ModInt &x) const { return ModInt()._setval((lint)val + x.val); }
MDCONST ModInt operator-(const ModInt &x) const {
return ModInt()._setval((lint)val - x.val + md);
}
MDCONST ModInt operator*(const ModInt &x) const {
return ModInt()._setval((lint)val * x.val % md);
}
MDCONST ModInt operator/(const ModInt &x) const {
return ModInt()._setval((lint)val * x.inv() % md);
}
MDCONST ModInt operator-() const { return ModInt()._setval(md - val); }
MDCONST ModInt &operator+=(const ModInt &x) { return *this = *this + x; }
MDCONST ModInt &operator-=(const ModInt &x) { return *this = *this - x; }
MDCONST ModInt &operator*=(const ModInt &x) { return *this = *this * x; }
MDCONST ModInt &operator/=(const ModInt &x) { return *this = *this / x; }
friend MDCONST ModInt operator+(lint a, const ModInt &x) {
return ModInt()._setval(a % md + x.val);
}
friend MDCONST ModInt operator-(lint a, const ModInt &x) {
return ModInt()._setval(a % md - x.val + md);
}
friend MDCONST ModInt operator*(lint a, const ModInt &x) {
return ModInt()._setval(a % md * x.val % md);
}
friend MDCONST ModInt operator/(lint a, const ModInt &x) {
return ModInt()._setval(a % md * x.inv() % md);
}
MDCONST bool operator==(const ModInt &x) const { return val == x.val; }
MDCONST bool operator!=(const ModInt &x) const { return val != x.val; }
MDCONST bool operator<(const ModInt &x) const {
return val < x.val;
} // To use std::map<ModInt, T>
friend std::istream &operator>>(std::istream &is, ModInt &x) {
lint t;
return is >> t, x = ModInt(t), is;
}
MDCONST friend std::ostream &operator<<(std::ostream &os, const ModInt &x) {
return os << x.val;
}
MDCONST ModInt pow(lint n) const {
ModInt ans = 1, tmp = *this;
while (n) {
if (n & 1) ans *= tmp;
tmp *= tmp, n >>= 1;
}
return ans;
}

static std::vector<ModInt> facs, facinvs, invs;
MDCONST static void _precalculation(int N) {
int l0 = facs.size();
if (N > md) N = md;
if (N <= l0) return;
facs.resize(N), facinvs.resize(N), invs.resize(N);
for (int i = l0; i < N; i++) facs[i] = facs[i - 1] * i;
facinvs[N - 1] = facs.back().pow(md - 2);
for (int i = N - 2; i >= l0; i--) facinvs[i] = facinvs[i + 1] * (i + 1);
for (int i = N - 1; i >= l0; i--) invs[i] = facinvs[i] * facs[i - 1];
}
MDCONST lint inv() const {
if (this->val < std::min(md >> 1, 1 << 21)) {
while (this->val >= int(facs.size())) _precalculation(facs.size() * 2);
return invs[this->val].val;
} else {
return this->pow(md - 2).val;
}
}
MDCONST ModInt fac() const {
while (this->val >= int(facs.size())) _precalculation(facs.size() * 2);
return facs[this->val];
}
MDCONST ModInt facinv() const {
while (this->val >= int(facs.size())) _precalculation(facs.size() * 2);
return facinvs[this->val];
}
MDCONST ModInt doublefac() const {
lint k = (this->val + 1) / 2;
return (this->val & 1) ? ModInt(k * 2).fac() / (ModInt(2).pow(k) * ModInt(k).fac())
: ModInt(k).fac() * ModInt(2).pow(k);
}
MDCONST ModInt nCr(const ModInt &r) const {
return (this->val < r.val) ? 0 : this->fac() * (*this - r).facinv() * r.facinv();
}
MDCONST ModInt nPr(const ModInt &r) const {
return (this->val < r.val) ? 0 : this->fac() * (*this - r).facinv();
}

ModInt sqrt() const {
if (val == 0) return 0;
if (md == 2) return val;
if (pow((md - 1) / 2) != 1) return 0;
ModInt b = 1;
while (b.pow((md - 1) / 2) == 1) b += 1;
int e = 0, m = md - 1;
while (m % 2 == 0) m >>= 1, e++;
ModInt x = pow((m - 1) / 2), y = (*this) * x * x;
x *= (*this);
ModInt z = b.pow(m);
while (y != 1) {
int j = 0;
ModInt t = y;
while (t != 1) j++, t *= t;
z = z.pow(1LL << (e - j - 1));
x *= z, z *= z, y *= z;
e = j;
}
return ModInt(std::min(x.val, md - x.val));
}
};
template <int md> std::vector<ModInt<md>> ModInt<md>::facs = {1};
template <int md> std::vector<ModInt<md>> ModInt<md>::facinvs = {1};
template <int md> std::vector<ModInt<md>> ModInt<md>::invs = {0};
using mint = ModInt<998244353>;

// Linear sieve algorithm for fast prime factorization
// Complexity: O(N) time, O(N) space:
// - MAXN = 10^7:  ~44 MB,  80~100 ms (Codeforces / AtCoder GCC, C++17)
// - MAXN = 10^8: ~435 MB, 810~980 ms (Codeforces / AtCoder GCC, C++17)
// Reference:
// [1] D. Gries, J. Misra, "A Linear Sieve Algorithm for Finding Prime Numbers,"
//     Communications of the ACM, 21(12), 999-1003, 1978.
// - https://cp-algorithms.com/algebra/prime-sieve-linear.html
// - https://37zigen.com/linear-sieve/
struct Sieve {
std::vector<int> min_factor;
std::vector<int> primes;
Sieve(int MAXN) : min_factor(MAXN + 1) {
for (int d = 2; d <= MAXN; d++) {
if (!min_factor[d]) {
min_factor[d] = d;
primes.emplace_back(d);
}
for (const auto &p : primes) {
if (p > min_factor[d] or d * p > MAXN) break;
min_factor[d * p] = p;
}
}
}
// Prime factorization for 1 <= x <= MAXN^2
// Complexity: O(log x)           (x <= MAXN)
//             O(MAXN / log MAXN) (MAXN < x <= MAXN^2)
template <class T> std::map<T, int> factorize(T x) const {
std::map<T, int> ret;
assert(x > 0 and
x <= ((long long)min_factor.size() - 1) * ((long long)min_factor.size() - 1));
for (const auto &p : primes) {
if (x < T(min_factor.size())) break;
while (!(x % p)) x /= p, ret[p]++;
}
if (x >= T(min_factor.size())) ret[x]++, x = 1;
while (x > 1) ret[min_factor[x]]++, x /= min_factor[x];
return ret;
}
// Enumerate divisors of 1 <= x <= MAXN^2
// Be careful of highly composite numbers https://oeis.org/A002182/list
// https://gist.github.com/dario2994/fb4713f252ca86c1254d#file-list-txt (n, (# of div. of n)):
// 45360->100, 735134400(<1e9)->1344, 963761198400(<1e12)->6720
template <class T> std::vector<T> divisors(T x) const {
std::vector<T> ret{1};
for (const auto p : factorize(x)) {
int n = ret.size();
for (int i = 0; i < n; i++) {
for (T a = 1, d = 1; d <= p.second; d++) {
a *= p.first;
ret.push_back(ret[i] * a);
}
}
}
return ret; // NOT sorted
}
// Euler phi functions of divisors of given x
// Complexity: O(sqrt(x) + d(x))
template <class T> std::map<T, T> euler_of_divisors(T x) const {
assert(x >= 1);
std::map<T, T> ret;
ret[1] = 1;
std::vector<T> divs{1};
for (auto p : factorize(x)) {
int n = ret.size();
for (int i = 0; i < n; i++) {
ret[divs[i] * p.first] = ret[divs[i]] * (p.first - 1);
divs.push_back(divs[i] * p.first);
for (T a = divs[i] * p.first, d = 1; d < p.second; a *= p.first, d++) {
ret[a * p.first] = ret[a] * p.first;
divs.push_back(a * p.first);
}
}
}
return ret;
}
// Moebius function Table, (-1)^{# of different prime factors} for square-free x
// return: [0=>0, 1=>1, 2=>-1, 3=>-1, 4=>0, 5=>-1, 6=>1, 7=>-1, 8=>0, ...] https://oeis.org/A008683
std::vector<int> GenerateMoebiusFunctionTable() const {
std::vector<int> ret(min_factor.size());
for (unsigned i = 1; i < min_factor.size(); i++) {
if (i == 1) {
ret[i] = 1;
} else if ((i / min_factor[i]) % min_factor[i] == 0) {
ret[i] = 0;
} else {
ret[i] = -ret[i / min_factor[i]];
}
}
return ret;
}
// Calculate [0^K, 1^K, ..., nmax^K] in O(nmax)
// Note: **0^0 == 1**
template <class MODINT> std::vector<MODINT> enumerate_kth_pows(long long K, int nmax) const {
assert(nmax < int(min_factor.size()));
assert(K >= 0);
if (K == 0) return std::vector<MODINT>(nmax + 1, 1);
std::vector<MODINT> ret(nmax + 1);
ret[0] = 0, ret[1] = 1;
for (int n = 2; n <= nmax; n++) {
if (min_factor[n] == n) {
ret[n] = MODINT(n).pow(K);
} else {
ret[n] = ret[n / min_factor[n]] * ret[min_factor[n]];
}
}
return ret;
}
};

// Integer convolution for arbitrary mod
// with NTT (and Garner's algorithm) for ModInt / ModIntRuntime class.
// We skip Garner's algorithm if skip_garner is true or mod is in nttprimes.
// input: a (size: n), b (size: m)
// return: vector (size: n + m - 1)
template <typename MODINT>
std::vector<MODINT> nttconv(std::vector<MODINT> a, std::vector<MODINT> b, bool skip_garner);

constexpr int nttprimes[3] = {998244353, 167772161, 469762049};

// Integer FFT (Fast Fourier Transform) for ModInt class
// (Also known as Number Theoretic Transform, NTT)
// is_inverse: inverse transform
// ** Input size must be 2^n **
template <typename MODINT> void ntt(std::vector<MODINT> &a, bool is_inverse = false) {
int n = a.size();
if (n == 1) return;
static const int mod = MODINT::mod();
static const MODINT root = MODINT::get_primitive_root();
assert(__builtin_popcount(n) == 1 and (mod - 1) % n == 0);

static std::vector<MODINT> w{1}, iw{1};
for (int m = w.size(); m < n / 2; m *= 2) {
MODINT dw = root.pow((mod - 1) / (4 * m)), dwinv = 1 / dw;
w.resize(m * 2), iw.resize(m * 2);
for (int i = 0; i < m; i++) w[m + i] = w[i] * dw, iw[m + i] = iw[i] * dwinv;
}

if (!is_inverse) {
for (int m = n; m >>= 1;) {
for (int s = 0, k = 0; s < n; s += 2 * m, k++) {
for (int i = s; i < s + m; i++) {
MODINT x = a[i], y = a[i + m] * w[k];
a[i] = x + y, a[i + m] = x - y;
}
}
}
} else {
for (int m = 1; m < n; m *= 2) {
for (int s = 0, k = 0; s < n; s += 2 * m, k++) {
for (int i = s; i < s + m; i++) {
MODINT x = a[i], y = a[i + m];
a[i] = x + y, a[i + m] = (x - y) * iw[k];
}
}
}
int n_inv = MODINT(n).inv();
for (auto &v : a) v *= n_inv;
}
}
template <int MOD>
std::vector<ModInt<MOD>> nttconv_(const std::vector<int> &a, const std::vector<int> &b) {
int sz = a.size();
assert(a.size() == b.size() and __builtin_popcount(sz) == 1);
std::vector<ModInt<MOD>> ap(sz), bp(sz);
for (int i = 0; i < sz; i++) ap[i] = a[i], bp[i] = b[i];
ntt(ap, false);
if (a == b)
bp = ap;
else
ntt(bp, false);
for (int i = 0; i < sz; i++) ap[i] *= bp[i];
ntt(ap, true);
return ap;
}
long long garner_ntt_(int r0, int r1, int r2, int mod) {
using mint2 = ModInt<nttprimes[2]>;
static const long long m01 = 1LL * nttprimes[0] * nttprimes[1];
static const long long m0_inv_m1 = ModInt<nttprimes[1]>(nttprimes[0]).inv();
static const long long m01_inv_m2 = mint2(m01).inv();

int v1 = (m0_inv_m1 * (r1 + nttprimes[1] - r0)) % nttprimes[1];
auto v2 = (mint2(r2) - r0 - mint2(nttprimes[0]) * v1) * m01_inv_m2;
return (r0 + 1LL * nttprimes[0] * v1 + m01 % mod * v2.val) % mod;
}
template <typename MODINT>
std::vector<MODINT> nttconv(std::vector<MODINT> a, std::vector<MODINT> b, bool skip_garner) {
if (a.empty() or b.empty()) return {};
int sz = 1, n = a.size(), m = b.size();
while (sz < n + m) sz <<= 1;
if (sz <= 16) {
std::vector<MODINT> ret(n + m - 1);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) ret[i + j] += a[i] * b[j];
}
return ret;
}
int mod = MODINT::mod();
if (skip_garner or
std::find(std::begin(nttprimes), std::end(nttprimes), mod) != std::end(nttprimes)) {
a.resize(sz), b.resize(sz);
if (a == b) {
ntt(a, false);
b = a;
} else {
ntt(a, false), ntt(b, false);
}
for (int i = 0; i < sz; i++) a[i] *= b[i];
ntt(a, true);
a.resize(n + m - 1);
} else {
std::vector<int> ai(sz), bi(sz);
for (int i = 0; i < n; i++) ai[i] = a[i].val;
for (int i = 0; i < m; i++) bi[i] = b[i].val;
auto ntt0 = nttconv_<nttprimes[0]>(ai, bi);
auto ntt1 = nttconv_<nttprimes[1]>(ai, bi);
auto ntt2 = nttconv_<nttprimes[2]>(ai, bi);
a.resize(n + m - 1);
for (int i = 0; i < n + m - 1; i++)
a[i] = garner_ntt_(ntt0[i].val, ntt1[i].val, ntt2[i].val, mod);
}
return a;
}

template <typename MODINT>
std::vector<MODINT> nttconv(const std::vector<MODINT> &a, const std::vector<MODINT> &b) {
return nttconv<MODINT>(a, b, false);
}

// Convert factorial power -> sampling
// [y[0], y[1], ..., y[N - 1]] -> \sum_i a_i x^\underline{i}
// Complexity: O(N log N)
template <class T> std::vector<T> factorial_to_ys(const std::vector<T> &as) {
const int N = as.size();
std::vector<T> exp(N, 1);
for (int i = 1; i < N; i++) exp[i] = T(i).facinv();
auto ys = nttconv(as, exp);
ys.resize(N);
for (int i = 0; i < N; i++) ys[i] *= T(i).fac();
return ys;
}

// Convert sampling -> factorial power
// [y[0], y[1], ..., y[N - 1]] -> \sum_i a_i x^\underline{i}
// Complexity: O(N log N)
template <class T> std::vector<T> ys_to_factorial(std::vector<T> ys) {
const int N = ys.size();
for (int i = 1; i < N; i++) ys[i] *= T(i).facinv();
std::vector<T> expinv(N, 1);
for (int i = 1; i < N; i++) expinv[i] = T(i).facinv() * (i % 2 ? -1 : 1);
auto as = nttconv(ys, expinv);
as.resize(N);
return as;
}

// Compute factorial power series of f(x + shift) from that of f(x)
// Complexity: O(N log N)
template <class T> std::vector<T> shift_of_factorial(const std::vector<T> &as, T shift) {
const int N = as.size();
std::vector<T> b(N, 1), c(N, 1);
for (int i = 1; i < N; i++) b[i] = b[i - 1] * (shift - i + 1) * T(i).inv();
for (int i = 0; i < N; i++) c[i] = as[i] * T(i).fac();
std::reverse(c.begin(), c.end());
auto ret = nttconv(b, c);
ret.resize(N);
std::reverse(ret.begin(), ret.end());
for (int i = 0; i < N; i++) ret[i] *= T(i).facinv();
return ret;
}

// Compute y[c], ..., y[c + width - 1] from y[0], ..., y[N - 1] where y = f(x) is poly of up to
// (N - 1)-th degree Complexity: O(N log N)
template <class T>
std::vector<T> shift_of_sampling_points(const std::vector<T> ys, T c, int width) {
auto as = ys_to_factorial(ys);
as = shift_of_factorial(as, c);
as.resize(width);
auto ret = factorial_to_ys(as);
return ret;
}

mint a272644(int n, int m) {
auto ys = Sieve(m + 1).enumerate_kth_pows<mint>(m + 1, m + 1);
const auto stirling_number_of_2nd = ys_to_factorial(ys);
mint ret = 0;
REP(i, m + 1) {
mint c = stirling_number_of_2nd[i + 1] * mint(i).pow(n - m) * mint(i).fac();
if ((m - i) & 1) ret -= c;
else {
ret += c;
}
}
return ret;
}

mint bruteforce(int N, int A) {
vector<int> is(N);
REP(i, N) is[i] = i;
int ret = 0;
do {
bool good = true;
REP(i, N) {
if (is[i] >= i + A or is[i] <= i - N + A) good = false;
}
ret += good;
} while (next_permutation(ALL(is)));
return ret;
}

int main() {

// FOR(n, 1, 10) {
//     vector<mint> sol;
//     FOR(k, 1, n) {
//         sol.push_back(bruteforce(n, k));
//     }
//     dbg(sol);
// }
// FOR(n, 1, 10) {
//     vector<mint> sol;
//     FOR(k, 1, n) {
//         sol.push_back(a272644(n, k));
//     }
//     dbg(sol);
// }

int N, A;
cin >> N >> A;
cout << a272644(N, A) << '\n';
}

##### Finally, Non Ad-Hoc Problem CodeChef Solution Review:

In our experience, we suggest you solve this Finally, Non Ad-Hoc Problem CodeChef Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

Non Ad-Hoc Problem Problem is available on Hacker Rank for Free, if you are stuck anywhere between a compilation, just visit Queslers to get Finally, Non Ad-Hoc Problem CodeChef Solution.

##### Conclusion:

I hope this Finally, Non Ad-Hoc Problem CodeChef Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Hacker Rank, Leetcode, Codechef, Codeforce Solution.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More CodeChef Solutions >>

Olympics Ranking CodeChef Solution

Problem Difficulties CodeChef Solution

Chef and Bulb Invention CodeChef Solution

Array Filling CodeChef Solution

Special Triplets CodeChef Solution