#include <bits/stdc++.h>
using namespace std;
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (auto &i : a) in >> i;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, vector<T> &a) {
  for (auto &i : a) out << i << ' ';
  return out;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, pair<T, U> &a) {
  out << a.first << ' ' << a.second << "\n";
  return out;
}
mt19937 rnd(time(NULL));
void files() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
template <typename T1, typename T2>
inline T1 min(T1 a, T2 b) {
  return (a < b ? a : b);
}
template <typename T1, typename T2>
inline T1 max(T1 a, T2 b) {
  return (a > b ? a : b);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  long long second = accumulate((a).begin(), (a).end(), (long long)0);
  if (second % n)
    cout << 1 << "\n";
  else
    cout << 0 << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q = 1;
  cin >> q;
  while (q--) {
    solve();
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (auto &i : a) in >> i;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, vector<T> &a) {
  for (auto &i : a) out << i << ' ';
  return out;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, pair<T, U> &a) {
  out << a.first << ' ' << a.second << "\n";
  return out;
}
mt19937 rnd(time(NULL));
void files() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
template <typename T1, typename T2>
inline T1 min(T1 a, T2 b) {
  return (a < b ? a : b);
}
template <typename T1, typename T2>
inline T1 max(T1 a, T2 b) {
  return (a > b ? a : b);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  long long second = accumulate((a).begin(), (a).end(), (long long)0);
  if (second % n)
    cout << 1 << "\n";
  else
    cout << 0 << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q = 1;
  cin >> q;
  while (q--) {
    solve();
  }
}

```