#include <bits/stdc++.h>
using namespace std;
void Err() {}
template <class T, class... A>
void Err(T a, A... x) {
  cerr << a << ',';
  Err(x...);
}
template <class X, class Y, class... A>
void Err(pair<X, Y> a, A... x) {
  cerr << '(' << a.first << ',' << a.second << "),";
  Err(x...);
}
template <template <class...> class T, class t, class... A>
void Err(T<t> a, A... x) {
  cerr << a.size() << ":{";
  for (auto v : a) Err(v);
  cerr << "},";
  Err(x...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rnd(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
template <class T>
void Min(T &a, const T b) {
  if (a > b) a = b;
}
template <class T>
void Max(T &a, const T b) {
  if (a < b) a = b;
}
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int N = 5e5 + 10, M = 500;
const long long mod = 1e9 + 7;
const long double eps = 1e-8;
int sgn(long double a) { return a < -eps ? -1 : (a < eps ? 0 : 1); }
int solve() {
  int n;
  cin >> n;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    sum += a;
  }
  if (sum % n == 0)
    return 0;
  else
    return 1;
}
int main() {
  ;
  ios::sync_with_stdio(0), cin.tie(0);
  int _;
  cin >> _;
  while (_--) {
    cout << (solve()) << '\n';
  }
}

### Prompt

Generate a CPP solution to 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;
void Err() {}
template <class T, class... A>
void Err(T a, A... x) {
  cerr << a << ',';
  Err(x...);
}
template <class X, class Y, class... A>
void Err(pair<X, Y> a, A... x) {
  cerr << '(' << a.first << ',' << a.second << "),";
  Err(x...);
}
template <template <class...> class T, class t, class... A>
void Err(T<t> a, A... x) {
  cerr << a.size() << ":{";
  for (auto v : a) Err(v);
  cerr << "},";
  Err(x...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rnd(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
template <class T>
void Min(T &a, const T b) {
  if (a > b) a = b;
}
template <class T>
void Max(T &a, const T b) {
  if (a < b) a = b;
}
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int N = 5e5 + 10, M = 500;
const long long mod = 1e9 + 7;
const long double eps = 1e-8;
int sgn(long double a) { return a < -eps ? -1 : (a < eps ? 0 : 1); }
int solve() {
  int n;
  cin >> n;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    sum += a;
  }
  if (sum % n == 0)
    return 0;
  else
    return 1;
}
int main() {
  ;
  ios::sync_with_stdio(0), cin.tie(0);
  int _;
  cin >> _;
  while (_--) {
    cout << (solve()) << '\n';
  }
}

```