#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;
}

### Prompt

Generate a Cpp solution to the following problem:

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.

### Response

```cpp
#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;
}

```