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You have prepared four problems. The difficulty levels of the problems are A1,A2,A3,A4A1,A2,A3,A4 respectively. A problem set comprises at least two problems and no two problems in a problem set should have the same difficulty level. A problem can belong to at most one problem set. Find the maximum number of problem sets you can create using the four problems.

- The first line of the input contains a single integer TT denoting the number of test cases. The description of TT test cases follows.
- The first and only line of each test case contains four space-separated integers A1A1, A2A2, A3A3, A4A4, denoting the difficulty level of four problems.

For each test case, print a single line containing one integer – the maximum number of problem sets you can create using the four problems.

- 1≤T≤10001≤T≤1000
- 1≤A1,A2,A3,A4≤101≤A1,A2,A3,A4≤10

**Subtask #1 (100 points):** Original constraints

```
3
1 4 3 2
4 5 5 5
2 2 2 2
```

```
2
1
0
```

**Test case 11:** You can prepare the first problem set using the first two problems and the second problem set using the next two problems. So the problem sets will be [1,4][1,4] and [3,2][3,2].

**Test case 22:** You can prepare one problem set using one problem having a difficulty level of 44 and the other having a difficulty level of 55. There is no way to prepare more than one problem set.

**Test case 33:** There is no way to prepare a problem set.

```
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
int main() {
// your code goes here
int t;
cin>>t;
while(t--)
{
int n=4;
int a[n];
for (int i = 0; i < n; i++) {
cin>>a[i];
}
sort(a,a+n);
if(a[0]==a[1] && a[1]==a[2] && a[2]==a[3])
cout<<"0"<<endl;
else if(a[0]!=a[1] && a[1]!=a[2] && a[2]!=a[3])
cout<<"2"<<endl;
else if(a[0]==a[1] && a[1]==a[2])
cout<<"1"<<endl;
else if(a[1]==a[2] && a[2]==a[3])
cout<<"1"<<endl;
else
cout<<"2"<<endl;
}
return 0;
}
```

```
t = int(input())
for _ in range(t):
inputt = [int(i) for i in input().split()]
inputt.sort()
if(inputt[0]==inputt[1] and inputt[2]==inputt[3] and inputt[0]!=inputt[3]):
print(2)
continue
inputt = set(inputt)
if len(inputt) == 4 or len(inputt) == 3:
print(2)
continue
if len(inputt) ==2:
print(1)
else:
print(0)
```

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