#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long sz = 2e6 + 5;
vector<long long> fac, ifac;
vector<int> prime;
long long mpow(long long a, long long b = mod - 2, long long m = mod) {
  long long res = 1;
  a %= m;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
  }
  return res;
}
void init() {
  fac.resize(sz, 1);
  ifac.resize(sz);
  for (long long i = 2; i < sz; i++) fac[i] = (fac[i - 1] * i) % mod;
  ifac[sz - 1] = mpow(fac[sz - 1]);
  for (long long i = sz - 2; i >= 0; i--)
    ifac[i] = (ifac[i + 1] * (i + 1)) % mod;
  prime.resize(sz);
  for (int i = 2; i < sz; i++) prime[i] = i;
  for (long long i = 2; i * i < sz; i++) {
    if (prime[i] != i) continue;
    for (long long j = i * i; j < sz; j += i)
      if (prime[j] == j) prime[j] = i;
  }
}
long long nCr(long long n, long long r) {
  if (n <= 0 || r < 0) return 0;
  if (n < r) return 0;
  return (fac[n] * (ifac[r] * ifac[n - r] % mod)) % mod;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
void solve() {
  long long n;
  cin >> n;
  long long sm = 0;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    sm += x;
  }
  if (sm % n == 0)
    cout << 0 << "\n";
  else
    cout << 1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  init();
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const long long mod = 998244353;
const long long sz = 2e6 + 5;
vector<long long> fac, ifac;
vector<int> prime;
long long mpow(long long a, long long b = mod - 2, long long m = mod) {
  long long res = 1;
  a %= m;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
  }
  return res;
}
void init() {
  fac.resize(sz, 1);
  ifac.resize(sz);
  for (long long i = 2; i < sz; i++) fac[i] = (fac[i - 1] * i) % mod;
  ifac[sz - 1] = mpow(fac[sz - 1]);
  for (long long i = sz - 2; i >= 0; i--)
    ifac[i] = (ifac[i + 1] * (i + 1)) % mod;
  prime.resize(sz);
  for (int i = 2; i < sz; i++) prime[i] = i;
  for (long long i = 2; i * i < sz; i++) {
    if (prime[i] != i) continue;
    for (long long j = i * i; j < sz; j += i)
      if (prime[j] == j) prime[j] = i;
  }
}
long long nCr(long long n, long long r) {
  if (n <= 0 || r < 0) return 0;
  if (n < r) return 0;
  return (fac[n] * (ifac[r] * ifac[n - r] % mod)) % mod;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
void solve() {
  long long n;
  cin >> n;
  long long sm = 0;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    sm += x;
  }
  if (sm % n == 0)
    cout << 0 << "\n";
  else
    cout << 1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  init();
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```