#include <bits/stdc++.h>
using namespace std;
const char ENDL = '\n';
const int MOD = 1000000007;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long n;
    cin >> n;
    vector<long long> ans;
    vector<long long> a;
    long long sum = 0;
    for (int j = 0; j < n; j++) {
      long long k;
      cin >> k;
      a.push_back(k);
      sum += k;
    }
    if (sum % (n * (n + 1) / 2) == 0) {
      sum /= n * (n + 1) / 2;
      bool flag = true;
      for (long long j = 0; j < n; j++) {
        if (j == 0) {
          if ((a[n - 1] - a[j] + sum) > 0 and
              (a[n - 1] - a[j] + sum) % n == 0) {
            ans.push_back((a[n - 1] - a[j] + sum) / n);
          } else {
            flag = false;
            break;
          }
        } else {
          if ((a[j - 1] - a[j] + sum) > 0 and
              (a[j - 1] - a[j] + sum) % n == 0) {
            ans.push_back((a[j - 1] - a[j] + sum) / n);
          } else {
            flag = false;
            break;
          }
        }
      }
      if (flag) {
        cout << "YES" << ENDL;
        for (long long j = 0; j < n; j++) {
          cout << ans[j] << ' ';
        }
        cout << ENDL;
      } else {
        cout << "NO" << ENDL;
      }
    } else {
      cout << "NO" << ENDL;
    }
  }
  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;
const char ENDL = '\n';
const int MOD = 1000000007;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long n;
    cin >> n;
    vector<long long> ans;
    vector<long long> a;
    long long sum = 0;
    for (int j = 0; j < n; j++) {
      long long k;
      cin >> k;
      a.push_back(k);
      sum += k;
    }
    if (sum % (n * (n + 1) / 2) == 0) {
      sum /= n * (n + 1) / 2;
      bool flag = true;
      for (long long j = 0; j < n; j++) {
        if (j == 0) {
          if ((a[n - 1] - a[j] + sum) > 0 and
              (a[n - 1] - a[j] + sum) % n == 0) {
            ans.push_back((a[n - 1] - a[j] + sum) / n);
          } else {
            flag = false;
            break;
          }
        } else {
          if ((a[j - 1] - a[j] + sum) > 0 and
              (a[j - 1] - a[j] + sum) % n == 0) {
            ans.push_back((a[j - 1] - a[j] + sum) / n);
          } else {
            flag = false;
            break;
          }
        }
      }
      if (flag) {
        cout << "YES" << ENDL;
        for (long long j = 0; j < n; j++) {
          cout << ans[j] << ' ';
        }
        cout << ENDL;
      } else {
        cout << "NO" << ENDL;
      }
    } else {
      cout << "NO" << ENDL;
    }
  }
  return 0;
}

```