#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int ans = power(a, b / 2);
  if (b % 2 == 0) return (ans * ans) % 1000000007;
  return ((ans * ans) % 1000000007 * a) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n, i;
    cin >> n;
    long long int b[n], sumb = 0, mn = INT_MAX, a[n];
    for (i = 0; i < n; i++) {
      cin >> b[i];
      mn = min(b[i], mn);
      sumb += b[i];
    }
    long long int p = n * (n + 1) / 2;
    if (sumb % p != 0) {
      cout << "NO\n";
    } else {
      long long int q = sumb / p;
      if (q < n) {
        cout << "NO\n";
      } else {
        long long int f = 1;
        for (i = 0; i < n; i++) {
          long long int d = 0;
          if (i == 0) {
            d = b[0] - b[n - 1];
            long long int r = q - d;
            if (r % n != 0) {
              f = 0;
              break;
            } else {
              a[0] = r / n;
            }
          } else {
            d = b[i] - b[i - 1];
            long long int r = q - d;
            if (r % n != 0) {
              f = 0;
              break;
            } else {
              a[i] = r / n;
            }
          }
          if (a[i] < 1) {
            f = 0;
            break;
          }
        }
        if (f == 0) {
          cout << "NO\n";
        } else {
          cout << "YES\n";
          for (i = 0; i < n; i++) {
            cout << a[i] << " ";
          }
          cout << "\n";
        }
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

n towns are arranged in a circle sequentially. The towns are numbered from 1 to n in clockwise order. In the i-th town, there lives a singer with a repertoire of a_i minutes for each i ∈ [1, n].

Each singer visited all n towns in clockwise order, starting with the town he lives in, and gave exactly one concert in each town. In addition, in each town, the i-th singer got inspired and came up with a song that lasts a_i minutes. The song was added to his repertoire so that he could perform it in the rest of the cities.

Hence, for the i-th singer, the concert in the i-th town will last a_i minutes, in the (i + 1)-th town the concert will last 2 ⋅ a_i minutes, ..., in the ((i + k) mod n + 1)-th town the duration of the concert will be (k + 2) ⋅ a_i, ..., in the town ((i + n - 2) mod n + 1) — n ⋅ a_i minutes.

You are given an array of b integer numbers, where b_i is the total duration of concerts in the i-th town. Reconstruct any correct sequence of positive integers a or say that it is impossible.

Input

The first line contains one integer t (1 ≤ t ≤ 10^3) — the number of test cases. Then the test cases follow.

Each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^4) — the number of cities. The second line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^{9}) — the total duration of concerts in i-th city.

The sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, print the answer as follows:

If there is no suitable sequence a, print NO. Otherwise, on the first line print YES, on the next line print the sequence a_1, a_2, ..., a_n of n integers, where a_i (1 ≤ a_i ≤ 10^{9}) is the initial duration of repertoire of the i-th singer. If there are multiple answers, print any of them.

Example

Input


4
3
12 16 14
1
1
3
1 2 3
6
81 75 75 93 93 87


Output


YES
3 1 3 
YES
1 
NO
YES
5 5 4 1 4 5 

Note

Let's consider the 1-st test case of the example:

  1. the 1-st singer in the 1-st city will give a concert for 3 minutes, in the 2-nd — for 6 minutes, in the 3-rd — for 9 minutes; 
  2. the 2-nd singer in the 1-st city will give a concert for 3 minutes, in the 2-nd — for 1 minute, in the 3-rd - for 2 minutes; 
  3. the 3-rd singer in the 1-st city will give a concert for 6 minutes, in the 2-nd — for 9 minutes, in the 3-rd — for 3 minutes. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int ans = power(a, b / 2);
  if (b % 2 == 0) return (ans * ans) % 1000000007;
  return ((ans * ans) % 1000000007 * a) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n, i;
    cin >> n;
    long long int b[n], sumb = 0, mn = INT_MAX, a[n];
    for (i = 0; i < n; i++) {
      cin >> b[i];
      mn = min(b[i], mn);
      sumb += b[i];
    }
    long long int p = n * (n + 1) / 2;
    if (sumb % p != 0) {
      cout << "NO\n";
    } else {
      long long int q = sumb / p;
      if (q < n) {
        cout << "NO\n";
      } else {
        long long int f = 1;
        for (i = 0; i < n; i++) {
          long long int d = 0;
          if (i == 0) {
            d = b[0] - b[n - 1];
            long long int r = q - d;
            if (r % n != 0) {
              f = 0;
              break;
            } else {
              a[0] = r / n;
            }
          } else {
            d = b[i] - b[i - 1];
            long long int r = q - d;
            if (r % n != 0) {
              f = 0;
              break;
            } else {
              a[i] = r / n;
            }
          }
          if (a[i] < 1) {
            f = 0;
            break;
          }
        }
        if (f == 0) {
          cout << "NO\n";
        } else {
          cout << "YES\n";
          for (i = 0; i < n; i++) {
            cout << a[i] << " ";
          }
          cout << "\n";
        }
      }
    }
  }
  return 0;
}

```