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 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;
  }
  