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.
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
const long long int INF = 1e9;
vector<long long int> sieve(long long int n) {
  int* arr = new int[n + 1]();
  vector<long long int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long int modMul(long long int x, long long int y) {
  x %= mod, y %= mod;
  return (x * 1ll * y) % mod;
}
long long int powM(long long int x, long long int y) {
  if (y == 0) {
    return 1;
  } else if (y % 2 == 0) {
    long long int _tmp = powM(x, y / 2);
    return modMul(_tmp, _tmp);
  } else {
    return modMul(x, powM(x, y - 1));
  }
}
long long int binary_search(vector<long long int> v, long long int ans) {
  long long int lo = 0;
  long long int hi = v.size() - 1;
  long long int mid;
  while (lo <= hi) {
    mid = (hi + lo) / 2;
    if (v[mid] == ans) {
      return mid;
    } else if (v[mid] < ans) {
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return -1;
}
void solve() {
  long long int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    for (long long int i = 0; i < n; i++) {
      cin >> v[i];
    }
    sort((v).begin(), (v).end());
    for (int i = 0; i < n / 2; i++) {
      if (v[n - i - 1] - v[i] > 1) {
        v[i] = (v[i] + v[n - i - 1]) / 2;
        v[n - i - 1] = v[i];
      }
    }
    sort((v).begin(), (v).end());
    cout << v[n - 1] - v[0] << "\n";
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
