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;
long long int power(long long int n, long long int k) {
  long long int j = 1;
  if (k == 0) return 1;
  while (k > 0) {
    j = (j * n) % 1000000007;
    k--;
  }
  return j;
}
void SieveOfEratosthenes(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  prime[1] = false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t, i, j, l, n, k, p, p1;
  cin >> t;
  while (t--) {
    cin >> n;
    long long int a[n];
    for (i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    p = 0;
    long long int d = a[0] + a[n - 1];
    if (d % 2)
      p = 1;
    else {
      for (i = 0; i < n / 2; i++) {
        if (a[i] + a[n - i - 1] != d) {
          p = 1;
          break;
        }
      }
    }
    if (p)
      cout << 1;
    else
      cout << 0;
    cout << "\n";
  }
  return 0;
}
