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;
  inline long long read() {
    long long x = 0, f = 1;
    char c;
    for (c = getchar(); c < '0' || c > '9';
         f = ((c == '-') ? -1 : f), c = getchar())
      ;
    for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = getchar())
      ;
    return x * f;
  }
  template <typename T>
  inline void write(T a) {
    if (a == 0) {
      putchar('0');
      return;
    }
    if (a < 0) putchar('-'), a = -a;
    char c1[120];
    int h = 0;
    while (a) c1[++h] = a % 10 + '0', a /= 10;
    for (int i = h; i >= 1; --i) putchar(c1[i]);
  }
  inline void write_() { return; }
  template <typename T, typename... Args>
  inline void write_(T a, Args... b) {
    write(a);
    putchar(' ');
    write_(b...);
  }
  inline void writeln() {
    putchar('\n');
    return;
  }
  template <typename T, typename... Args>
  inline void writeln(T a, Args... b) {
    write(a);
    putchar(' ');
    writeln(b...);
  }
  long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
  inline long long lcm(long long a, long long b) {
    return 1ll * a / gcd(a, b) * b;
  }
  inline long long Pow(long long n, long long a) {
    long long b = 1;
    while (a) {
      if (a & 1) b = 1ll * b * n % (int)(1e9 + 7);
      n = 1ll * n * n % (int)(1e9 + 7);
      a >>= 1;
    }
    return b;
  }
  signed main() {
    int t = read(), x, n;
    long long sum = 0;
    while (t--) {
      n = read();
      sum = 0;
      for (int i = 1; i <= n; ++i) x = read(), sum += x;
      if (sum % n == 0)
        writeln(0);
      else
        writeln(1);
    }
    return 0;
  }
  