description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  using namespace chrono;
  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 << "]";
  }
  void solve() {
    long long n, sum = 0;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    for (long long i = 0; i < n; i++) sum += a[i];
    ;
    (sum % n == 0) ? cout << 0 : cout << 1;
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    auto start1 = high_resolution_clock::now();
    int test = 1;
    cin >> test;
    for (long long i = 1; i < test + 1; i++) {
      solve();
    }
    auto stop1 = high_resolution_clock::now();
    auto duration = duration_cast<nanoseconds>(stop1 - start1);
    cerr << "\n[Completed in "
         << ((long double)duration.count()) / ((long double)1e9) << "s]" << endl;
    return 0;
  }
  