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;
  const double pi = acos(-1.0);
  template <typename T, typename U>
  inline void amin(T &x, U y) {
    if (y < x) x = y;
  }
  template <typename T, typename U>
  inline void amax(T &x, U y) {
    if (x < y) x = y;
  }
  template <typename T>
  inline void write(T x) {
    int i = 20;
    char buf[21];
    buf[20] = '\n';
    do {
      buf[--i] = x % 10 + '0';
      x /= 10;
    } while (x);
    do {
      putchar(buf[i]);
    } while (buf[i++] != '\n');
  }
  template <typename T>
  inline T readInt() {
    T n = 0, s = 1;
    char p = getchar();
    if (p == '-') s = -1;
    while ((p < '0' || p > '9') && p != EOF && p != '-') p = getchar();
    if (p == '-') s = -1, p = getchar();
    while (p >= '0' && p <= '9') {
      n = (n << 3) + (n << 1) + (p - '0');
      p = getchar();
    }
    return n * s;
  }
  void i_am_just_a_regular_everyday_normal_MOFO() {
    long long int n, som = 0;
    n = readInt<long long int>();
    vector<int> a(n);
    for (long long int i = 0; i < n; i++) {
      a[i] = readInt<long long int>();
      som += a[i];
    }
    if (som % n == 0)
      cout << "0" << endl;
    else
      cout << "1" << endl;
  }
  signed main() {
    int t;
    t = readInt<int>();
    while (t--) {
      i_am_just_a_regular_everyday_normal_MOFO();
    }
    return 0;
  }
  