#include <bits/stdc++.h>
using namespace std;
bool sbs(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first > b.first;
}
bool isPrime(long long int n) {
  if (n == 2) return true;
  if (n < 2 || !(n & 1)) return false;
  for (long long int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
void xxxx(map<long long int, vector<pair<long long int, long long int>>> p,
          long long int l, vector<long long int> a, bool b[]) {
  b[l] = true;
  for (auto x : p[l]) {
    long long int y1 = x.first;
    long long int y2 = x.second;
    a[y1] = min(a[y1], a[l] + y2);
    xxxx(p, y1, a, b);
  }
}
void dfs(vector<long long int> &a,
         unordered_map<long long int, long long int> &m, long long int i,
         long long int k, long long int &res,
         unordered_map<long long int, bool> &visit) {
  if (visit.find(i) != visit.end()) return;
  if (a[i] == -1) {
    visit[i] = true;
    return;
  }
  if (m.find(i) != m.end()) {
    res = max(res, k - m[i]);
    visit[i] = true;
    return;
  }
  m[i] = k;
  dfs(a, m, a[i], k + 1, res, visit);
  visit[i] = true;
}
long long int nextPowerOf2(long long int n) {
  n--;
  n |= n >> 1;
  n |= n >> 2;
  n |= n >> 4;
  n |= n >> 8;
  n |= n >> 16;
  n++;
  return n;
}
void xxxx(map<long long int, vector<long long int>> &m, long long int s,
          bool visited[], long long int &k, vector<long long int> &f) {
  visited[s] = true;
  k++;
  f.push_back(s);
  for (auto u : m[s]) {
    if (visited[u] == false) {
      xxxx(m, u, visited, k, f);
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v;
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    v.push_back(x);
    m[x]++;
  }
  sort(v.begin(), v.end());
  long long int l = 0;
  long long int h = n - 1;
  while (true) {
    long long int l1 = INT_MAX;
    long long int l2 = INT_MIN;
    long long int k1 = -1;
    long long int k2 = -1;
    for (long long int i = 0; i < n; i++) {
      if (v[i] < l1) {
        l1 = v[i];
        k1 = i;
      }
      if (v[i] > l2) {
        l2 = v[i];
        k2 = i;
      }
    }
    if (l1 == l2 - 1 || l1 == l2 || k1 == k2) {
      break;
    } else {
      long long int k3 = l1 + l2;
      if (k3 % 2 == 0) {
        v[k1] = k3 / 2;
        v[k2] = k3 / 2;
      } else {
        v[k1] = k3 / 2;
        v[k2] = v[k1] + 1;
      }
    }
  }
  long long int m1 = INT_MIN;
  long long int m2 = INT_MAX;
  for (auto x : v) {
    m1 = max(m1, x);
    m2 = min(m2, x);
  }
  cout << m1 - m2 << endl;
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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;
bool sbs(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first > b.first;
}
bool isPrime(long long int n) {
  if (n == 2) return true;
  if (n < 2 || !(n & 1)) return false;
  for (long long int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
void xxxx(map<long long int, vector<pair<long long int, long long int>>> p,
          long long int l, vector<long long int> a, bool b[]) {
  b[l] = true;
  for (auto x : p[l]) {
    long long int y1 = x.first;
    long long int y2 = x.second;
    a[y1] = min(a[y1], a[l] + y2);
    xxxx(p, y1, a, b);
  }
}
void dfs(vector<long long int> &a,
         unordered_map<long long int, long long int> &m, long long int i,
         long long int k, long long int &res,
         unordered_map<long long int, bool> &visit) {
  if (visit.find(i) != visit.end()) return;
  if (a[i] == -1) {
    visit[i] = true;
    return;
  }
  if (m.find(i) != m.end()) {
    res = max(res, k - m[i]);
    visit[i] = true;
    return;
  }
  m[i] = k;
  dfs(a, m, a[i], k + 1, res, visit);
  visit[i] = true;
}
long long int nextPowerOf2(long long int n) {
  n--;
  n |= n >> 1;
  n |= n >> 2;
  n |= n >> 4;
  n |= n >> 8;
  n |= n >> 16;
  n++;
  return n;
}
void xxxx(map<long long int, vector<long long int>> &m, long long int s,
          bool visited[], long long int &k, vector<long long int> &f) {
  visited[s] = true;
  k++;
  f.push_back(s);
  for (auto u : m[s]) {
    if (visited[u] == false) {
      xxxx(m, u, visited, k, f);
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v;
  map<long long int, long long int> m;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    v.push_back(x);
    m[x]++;
  }
  sort(v.begin(), v.end());
  long long int l = 0;
  long long int h = n - 1;
  while (true) {
    long long int l1 = INT_MAX;
    long long int l2 = INT_MIN;
    long long int k1 = -1;
    long long int k2 = -1;
    for (long long int i = 0; i < n; i++) {
      if (v[i] < l1) {
        l1 = v[i];
        k1 = i;
      }
      if (v[i] > l2) {
        l2 = v[i];
        k2 = i;
      }
    }
    if (l1 == l2 - 1 || l1 == l2 || k1 == k2) {
      break;
    } else {
      long long int k3 = l1 + l2;
      if (k3 % 2 == 0) {
        v[k1] = k3 / 2;
        v[k2] = k3 / 2;
      } else {
        v[k1] = k3 / 2;
        v[k2] = v[k1] + 1;
      }
    }
  }
  long long int m1 = INT_MIN;
  long long int m2 = INT_MAX;
  for (auto x : v) {
    m1 = max(m1, x);
    m2 = min(m2, x);
  }
  cout << m1 - m2 << endl;
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```