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.
#include <bits/stdc++.h>
using namespace std;
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
  uniform_int_distribution<int> uid(0, lim - 1);
  return uid(rang);
}
const long long mod = 100000000007;
long long inv(long long i) {
  if (i == 1) return 1;
  return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long mod_mul(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a * b) % mod) + mod) % mod;
}
long long mod_add(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a + b) % mod) + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long ceil_div(long long a, long long b) {
  return a % b == 0 ? a / b : a / b + 1;
}
long long mpow(long long base, long long exp) {
  base %= mod;
  long long result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}
string DecimalToBinaryString(int a) {
  string binary = "";
  int mask = 1;
  for (int i = 0; i < 31; i++) {
    if ((mask & a) >= 1)
      binary = "1" + binary;
    else
      binary = "0" + binary;
    mask <<= 1;
  }
  cout << binary << endl;
  return binary;
}
string helper(string s, int k) {
  while (s.size() <= k) {
    s = s + s;
  }
  return s.substr(0, k);
}
void solve() {
  int n;
  cin >> n;
  long long a[n];
  int i;
  for (i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  vector<long long> ans;
  i = 1;
  int j = n;
  while (i <= j) {
    long long t1 =
        ((a[i] + a[j]) % 2 == 0) ? (a[i] + a[j]) / 2 : (a[i] + a[j]) / 2;
    long long t2 =
        ((a[i] + a[j]) % 2 == 0) ? (a[i] + a[j]) / 2 : (a[i] + a[j]) / 2 + 1;
    ans.push_back(t1);
    ans.push_back(t2);
    i++;
    j--;
  }
  sort(ans.begin(), ans.end());
  cout << ans[ans.size() - 1] - ans[0] << "\n";
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
