#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort((v).begin(), (v).end());
  long long sum = 0;
  long long mn = 1e7;
  for (long long i = 0; i < n; i++) {
    sum += v[i];
  }
  if (sum % n == 0) {
    cout << 0 << "\n";
  } else
    cout << 1 << "\n";
}
int32_t main() {
  cin.tie(0);
  cout.tie(NULL);
  ios_base::sync_with_stdio(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

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

There are n block towers in a row, where tower i has a height of a_i. You're part of a building crew, and you want to make the buildings look as nice as possible. In a single day, you can perform the following operation:

  * Choose two indices i and j (1 ≤ i, j ≤ n; i ≠ j), and move a block from tower i to tower j. This essentially decreases a_i by 1 and increases a_j by 1. 



You think the ugliness of the buildings is the height difference between the tallest and shortest buildings. Formally, the ugliness is defined as max(a)-min(a). 

What's the minimum possible ugliness you can achieve, after any number of days?

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t cases follow.

The first line of each test case contains one integer n (2 ≤ n ≤ 100) — the number of buildings.

The second line of each test case contains n space separated integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^7) — the heights of the buildings.

Output

For each test case, output a single integer — the minimum possible ugliness of the buildings.

Example

Input


3
3
10 10 10
4
3 2 1 2
5
1 2 3 1 5


Output


0
0
1

Note

In the first test case, the ugliness is already 0.

In the second test case, you should do one operation, with i = 1 and j = 3. The new heights will now be [2, 2, 2, 2], with an ugliness of 0.

In the third test case, you may do three operations: 

  1. with i = 3 and j = 1. The new array will now be [2, 2, 2, 1, 5], 
  2. with i = 5 and j = 4. The new array will now be [2, 2, 2, 2, 4], 
  3. with i = 5 and j = 3. The new array will now be [2, 2, 3, 2, 3]. 

The resulting ugliness is 1. It can be proven that this is the minimum possible ugliness for this test.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort((v).begin(), (v).end());
  long long sum = 0;
  long long mn = 1e7;
  for (long long i = 0; i < n; i++) {
    sum += v[i];
  }
  if (sum % n == 0) {
    cout << 0 << "\n";
  } else
    cout << 1 << "\n";
}
int32_t main() {
  cin.tie(0);
  cout.tie(NULL);
  ios_base::sync_with_stdio(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```