**Akash and Equal Mean solution codechef** – Akash goes to class everyday. His classes are held on the 9th9th floor so he uses a lift. One day NN people get on the lift. However, since the lift can carry at most N−2N−2 people, 22 of them need to get off.

## [Solution] Akash and Equal Mean solution codechef

Given an array AA of the weights of the NN people, Akash wonders how many ways of removing 22 people exist such that the mean of weights of the people in the lift remains constant. In other words, the mean of weights of the N−2N−2 people remaining in the lift must be the same as the mean of weights of all the NN people who were initially on the lift. Help Akash find how many such ways exist.

### Input Format

- The first line contains a single integer TT – the number of test cases. Then the test cases follow.
- The first line of each test case contains an integer NN – the number of people who get on the lift initially.
- The second line of each test case contains NN space-separated integers A1,A2,…,ANA1,A2,…,AN denoting the weights of the NN people on the lift.

### Output Format

For each test case, print a single integer denoting the number of ways of removing 22 people such that the mean reamins the same.

### Constraints

- 1≤T≤10001≤T≤1000
- 3≤N≤1053≤N≤105
- 1≤Ai≤1091≤Ai≤109
- Sum of NN over all test cases does not exceed 2⋅1052⋅105

## [Solution] Akash and Equal Mean solution codechef

```
3
4
1 3 5 7
4
1 7 7 7
6
1 1 1 1 1 1
```

### Sample Output 1

```
2
0
15
```

## Akash and Equal Mean solution Explanation

**Test case 11:** The mean of weights of all NN students is 1+3+5+74=164=41+3+5+74=164=4.

- Consider removing the people with weights (1,7)(1,7). The mean of weights of the N−2N−2 people remaining in the lift is 3+52=82=43+52=82=4, which is same as the mean of all NN people.
- Consider removing the people with weights (3,5)(3,5). The mean of weights of the N−2N−2 people remaining in the lift is 1+72=82=41+72=82=4, which is same as the mean of all NN people.

There is no other way to remove 22 people such that the mean reamins the same. Hence the answer is 22.

**Test case 22:** The mean of weights of all NN students is 1+7+7+74=5.51+7+7+74=5.5. It can be seen that there is no way to remove 22 people such that the mean reamins the same. Hence the answer is 00.