#include <bits/stdc++.h>
using namespace std;
long double PI = acosl(-1);
bool compare_int(int a, int b) { return (a > b); }
bool compare_string(string a, string b) { return a.size() < b.size(); }
bool compare_pair(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.second == b.second)
    return a.first < b.first;
  else
    return (a.second > b.second);
}
bool cmp(pair<string, int> x, pair<string, int> y) {
  return (x.second < y.second);
}
void NA() {
  printf("NO\n");
  exit(0);
}
void YA() {
  printf("YES\n");
  exit(0);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int x, tot = 0;
    for (int i = 0; i < n; i++) {
      cin >> x;
      tot += x;
    }
    tot % n == 0 ? cout << 0 : cout << 1;
    cout << "\n";
  }
}

### Prompt

Develop a solution in cpp 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;
long double PI = acosl(-1);
bool compare_int(int a, int b) { return (a > b); }
bool compare_string(string a, string b) { return a.size() < b.size(); }
bool compare_pair(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.second == b.second)
    return a.first < b.first;
  else
    return (a.second > b.second);
}
bool cmp(pair<string, int> x, pair<string, int> y) {
  return (x.second < y.second);
}
void NA() {
  printf("NO\n");
  exit(0);
}
void YA() {
  printf("YES\n");
  exit(0);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int x, tot = 0;
    for (int i = 0; i < n; i++) {
      cin >> x;
      tot += x;
    }
    tot % n == 0 ? cout << 0 : cout << 1;
    cout << "\n";
  }
}

```