#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(unordered_map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long expo(long long a, long long b, long long mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long expo(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b = b >> 1;
  }
  return res;
}
void extendgcd(long long a, long long b, long long *v) {
  if (b == 0) {
    v[0] = 1;
    v[1] = 0;
    v[2] = a;
    return;
  }
  extendgcd(b, a % b, v);
  long long x = v[1];
  v[1] = v[0] - v[1] * (a / b);
  v[0] = x;
  return;
}
long long mminv(long long a, long long b) {
  long long arr[3];
  extendgcd(a, b, arr);
  return arr[0];
}
long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
bool revsort(long long a, long long b) { return a > b; }
void swap(int &x, int &y) {
  int temp = x;
  x = y;
  y = temp;
}
long long combination(long long n, long long r, long long m, long long *fact,
                      long long *ifact) {
  long long val1 = fact[n];
  long long val2 = ifact[n - r];
  long long val3 = ifact[r];
  return (((val1 * val2) % m) * val3) % m;
}
void google(int t) { cout << "Case #" << t << ": "; }
vector<int> sieve(int n) {
  int *arr = new int[n + 1]();
  vector<int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (long long j = (long long(i) * long long(i)); j <= n; j += i)
        arr[j] = 1;
    }
  return vect;
}
long long phin(long long n) {
  long long number = n;
  if (n % 2 == 0) {
    number /= 2;
    while (n % 2 == 0) n /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      number = (number / i * (i - 1));
    }
  }
  if (n > 1) number = (number / n * (n - 1));
  return number;
}
long long inv(long long a) { return expo(a, mod - 2, mod); }
long long add(long long a, long long b) { return (a + b) % mod; }
long long sub(long long a, long long b) { return (a - b + mod) % mod; }
long long mul(long long a, long long b) { return (a * b) % mod; }
long long divi(long long a, long long b) { return (a * inv(b)) % mod; }
long long nCr(long long n, long long r) {
  if (r > n) return 0;
  r = min(r, n - r);
  vector<long long> dp(r + 1, 0);
  dp[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(r, i); j >= 1; j--) {
      dp[j] = (dp[j - 1] + dp[j] + mod) % mod;
    }
  }
  return dp[r];
}
long long sum2(long long a[], int n) {
  long long initial_sum = 0;
  return accumulate(a, a + n, initial_sum);
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long s = 0;
    for (long long(i) = 0; (i) < (long long)(n); ++(i)) {
      long long a;
      cin >> a;
      s += a;
    }
    cout << (s % n ? 1 : 0) << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

### Prompt

Please provide a Cpp coded solution 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;
const long long mod = 1e9 + 7;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(unordered_map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long expo(long long a, long long b, long long mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long expo(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b = b >> 1;
  }
  return res;
}
void extendgcd(long long a, long long b, long long *v) {
  if (b == 0) {
    v[0] = 1;
    v[1] = 0;
    v[2] = a;
    return;
  }
  extendgcd(b, a % b, v);
  long long x = v[1];
  v[1] = v[0] - v[1] * (a / b);
  v[0] = x;
  return;
}
long long mminv(long long a, long long b) {
  long long arr[3];
  extendgcd(a, b, arr);
  return arr[0];
}
long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
bool revsort(long long a, long long b) { return a > b; }
void swap(int &x, int &y) {
  int temp = x;
  x = y;
  y = temp;
}
long long combination(long long n, long long r, long long m, long long *fact,
                      long long *ifact) {
  long long val1 = fact[n];
  long long val2 = ifact[n - r];
  long long val3 = ifact[r];
  return (((val1 * val2) % m) * val3) % m;
}
void google(int t) { cout << "Case #" << t << ": "; }
vector<int> sieve(int n) {
  int *arr = new int[n + 1]();
  vector<int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (long long j = (long long(i) * long long(i)); j <= n; j += i)
        arr[j] = 1;
    }
  return vect;
}
long long phin(long long n) {
  long long number = n;
  if (n % 2 == 0) {
    number /= 2;
    while (n % 2 == 0) n /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      number = (number / i * (i - 1));
    }
  }
  if (n > 1) number = (number / n * (n - 1));
  return number;
}
long long inv(long long a) { return expo(a, mod - 2, mod); }
long long add(long long a, long long b) { return (a + b) % mod; }
long long sub(long long a, long long b) { return (a - b + mod) % mod; }
long long mul(long long a, long long b) { return (a * b) % mod; }
long long divi(long long a, long long b) { return (a * inv(b)) % mod; }
long long nCr(long long n, long long r) {
  if (r > n) return 0;
  r = min(r, n - r);
  vector<long long> dp(r + 1, 0);
  dp[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(r, i); j >= 1; j--) {
      dp[j] = (dp[j - 1] + dp[j] + mod) % mod;
    }
  }
  return dp[r];
}
long long sum2(long long a[], int n) {
  long long initial_sum = 0;
  return accumulate(a, a + n, initial_sum);
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long s = 0;
    for (long long(i) = 0; (i) < (long long)(n); ++(i)) {
      long long a;
      cin >> a;
      s += a;
    }
    cout << (s % n ? 1 : 0) << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

```