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You are given two **positive** integers `startPos`

and `endPos`

. Initially, you are standing at position `startPos`

on an **infinite** number line. With one step, you can move either one position to the left, or one position to the right.

Given a positive integer `k`

, return *the number of different ways to reach the position *

`endPos`

`startPos`

`k`

`10`^{9} + 7

.Two ways are considered different if the order of the steps made is not exactly the same.

**Note** that the number line includes negative integers.

**Example 1:**

**Input:** startPos = 1, endPos = 2, k = 3
**Output:** 3
**Explanation:** We can reach position 2 from 1 in exactly 3 steps in three ways:
- 1 -> 2 -> 3 -> 2.
- 1 -> 2 -> 1 -> 2.
- 1 -> 0 -> 1 -> 2.
It can be proven that no other way is possible, so we return 3.

**Example 2:**

**Input:** startPos = 2, endPos = 5, k = 10
**Output:** 0
**Explanation:** It is impossible to reach position 5 from position 2 in exactly 10 steps.

**Constraints:**

`1 <= startPos, endPos, k <= 1000`

```
int p = 1000000007;
public int numberOfWays(int a, int b, int k) {
if ((a - b - k) % 2 != 0) return 0;
if (Math.abs(a - b) > k) return 0;
long res = 1L;
for (int i = 0; i < (b - a + k) / 2; ++i) {
res = res * (k - i) % p;
res = res * inv(i + 1) % p;
}
return (int)res;
}
private long inv(long a) {
if (a == 1) return 1;
return (p - p / a) * inv(p % a) % p;
}
```

```
int p = 1e9 + 7;
int numberOfWays(int a, int b, int k) {
if ((a - b - k) % 2) return 0;
if (abs(a - b) > k) return 0;
long long res = 1L;
for (int i = 0; i < (b - a + k) / 2; ++i) {
res = res * (k - i) % p;
res = res * inv(i + 1) % p;
}
return res;
}
long inv(long a) {
if (a == 1) return 1;
return (p - p / a) * inv(p % a) % p;
}
```

```
def numberOfWays(self, a, b, k):
if (a - b - k) % 2: return 0
return comb(k, (b - a + k) // 2) % (10 ** 9 + 7)
```

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