#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
inline bool isSame(double c, double d) { return abs(c - d) < 1e-9; }
long long binpow(long long base, long long exp, int mod) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % mod;
    exp = exp >> 1;
    base = (base * base) % mod;
  }
  return res;
}
long long mod(long long x) {
  return ((x % 1000000007LL + 1000000007LL) % 1000000007LL);
}
long long add(long long a, long long b) { return mod(mod(a) + mod(b)); }
long long mul(long long a, long long b) { return mod(mod(a) * mod(b)); }
long long int decbin(string s) { return stoll(s, 0, 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int sum = 0;
    map<long long int, long long int> m;
    long long int arr[n];
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      sum += arr[i];
    }
    if (((2 * sum) % n) != 0) {
      cout << 0 << "\n";
    } else {
      long long int x = 0, t = (2 * sum) / n;
      for (int i = 0; i < n; i++) {
        x += m[t - arr[i]];
        m[arr[i]]++;
      }
      cout << x << "\n";
    }
  };
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

Monocarp has got an array a consisting of n integers. Let's denote k as the mathematic mean of these elements (note that it's possible that k is not an integer). 

The mathematic mean of an array of n elements is the sum of elements divided by the number of these elements (i. e. sum divided by n).

Monocarp wants to delete exactly two elements from a so that the mathematic mean of the remaining (n - 2) elements is still equal to k.

Your task is to calculate the number of pairs of positions [i, j] (i < j) such that if the elements on these positions are deleted, the mathematic mean of (n - 2) remaining elements is equal to k (that is, it is equal to the mathematic mean of n elements of the original array a).

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.

The first line of each testcase contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.

The second line contains a sequence of integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{9}), where a_i is the i-th element of the array.

The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.

Output

Print one integer — the number of pairs of positions [i, j] (i < j) such that if the elements on these positions are deleted, the mathematic mean of (n - 2) remaining elements is equal to k (that is, it is equal to the mathematic mean of n elements of the original array a).

Example

Input


4
4
8 8 8 8
3
50 20 10
5
1 4 7 3 5
7
1 2 3 4 5 6 7


Output


6
0
2
3

Note

In the first example, any pair of elements can be removed since all of them are equal.

In the second example, there is no way to delete two elements so the mathematic mean doesn't change.

In the third example, it is possible to delete the elements on positions 1 and 3, or the elements on positions 4 and 5.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
inline bool isSame(double c, double d) { return abs(c - d) < 1e-9; }
long long binpow(long long base, long long exp, int mod) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % mod;
    exp = exp >> 1;
    base = (base * base) % mod;
  }
  return res;
}
long long mod(long long x) {
  return ((x % 1000000007LL + 1000000007LL) % 1000000007LL);
}
long long add(long long a, long long b) { return mod(mod(a) + mod(b)); }
long long mul(long long a, long long b) { return mod(mod(a) * mod(b)); }
long long int decbin(string s) { return stoll(s, 0, 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int sum = 0;
    map<long long int, long long int> m;
    long long int arr[n];
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      sum += arr[i];
    }
    if (((2 * sum) % n) != 0) {
      cout << 0 << "\n";
    } else {
      long long int x = 0, t = (2 * sum) / n;
      for (int i = 0; i < n; i++) {
        x += m[t - arr[i]];
        m[arr[i]]++;
      }
      cout << x << "\n";
    }
  };
  return 0;
}

```