#include <bits/stdc++.h>
const int N = 1e5 + 10;
long long a[N], b[N];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n, res = 0, k = 0;
    scanf("%lld", &n);
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      k += a[i];
    }
    k *= 2;
    if (k % (n * (n + 1)) != 0) {
      printf("NO\n");
      continue;
    }
    k /= n * (n + 1);
    int L = n - 1;
    for (int i = 0; i < n; i++) {
      long long num = a[i] - k;
      if (num <= 0) {
        k = -1;
        break;
      }
      num = a[L % n] - num;
      if (num <= 0 || num % n != 0) {
        k = -1;
        break;
      }
      b[i] = num / n;
      L++;
    }
    if (n == 1) {
      b[0] = a[0];
      k = 0;
    }
    if (k == -1)
      printf("NO\n");
    else {
      printf("YES\n");
      for (int i = 0; i < n; i++) printf("%d ", b[i]);
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
const int N = 1e5 + 10;
long long a[N], b[N];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n, res = 0, k = 0;
    scanf("%lld", &n);
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      k += a[i];
    }
    k *= 2;
    if (k % (n * (n + 1)) != 0) {
      printf("NO\n");
      continue;
    }
    k /= n * (n + 1);
    int L = n - 1;
    for (int i = 0; i < n; i++) {
      long long num = a[i] - k;
      if (num <= 0) {
        k = -1;
        break;
      }
      num = a[L % n] - num;
      if (num <= 0 || num % n != 0) {
        k = -1;
        break;
      }
      b[i] = num / n;
      L++;
    }
    if (n == 1) {
      b[0] = a[0];
      k = 0;
    }
    if (k == -1)
      printf("NO\n");
    else {
      printf("YES\n");
      for (int i = 0; i < n; i++) printf("%d ", b[i]);
      printf("\n");
    }
  }
  return 0;
}

```