#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    bool state = true;
    long long a[n], sum = 0;
    vector<int> v;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    long long x = ((n + 1) * (n)) / 2;
    if (sum % x != 0) {
      state = false;
    } else {
      if (x == 0) {
        state = false;
      } else {
        sum = sum / x;
      }
    }
    long long zz = sum - (a[0] - a[n - 1]);
    if (zz % n != 0) {
      state = false;
    } else {
      zz = zz / n;
      v.push_back(zz);
      if (zz <= 0) {
        state = false;
      }
    }
    for (int i = 1; i < n; i++) {
      long long z = sum - (a[i] - a[i - 1]);
      if (z % n != 0) {
        state = false;
      } else {
        z = z / n;
        v.push_back(z);
        if (z <= 0) {
          state = false;
        }
      }
    }
    if (state) {
      long long newsum = 0;
      for (int i = 0; i < n; i++) {
        newsum += v[i];
      }
      if (newsum == sum) {
        cout << "YES"
             << "\n";
        for (int i = 0; i < n; i++) {
          cout << v[i] << " ";
        }
      } else {
        cout << "NO";
      }
      cout << "\n";
    } else {
      cout << "NO"
           << "\n";
    }
  }
}

### 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>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    bool state = true;
    long long a[n], sum = 0;
    vector<int> v;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    long long x = ((n + 1) * (n)) / 2;
    if (sum % x != 0) {
      state = false;
    } else {
      if (x == 0) {
        state = false;
      } else {
        sum = sum / x;
      }
    }
    long long zz = sum - (a[0] - a[n - 1]);
    if (zz % n != 0) {
      state = false;
    } else {
      zz = zz / n;
      v.push_back(zz);
      if (zz <= 0) {
        state = false;
      }
    }
    for (int i = 1; i < n; i++) {
      long long z = sum - (a[i] - a[i - 1]);
      if (z % n != 0) {
        state = false;
      } else {
        z = z / n;
        v.push_back(z);
        if (z <= 0) {
          state = false;
        }
      }
    }
    if (state) {
      long long newsum = 0;
      for (int i = 0; i < n; i++) {
        newsum += v[i];
      }
      if (newsum == sum) {
        cout << "YES"
             << "\n";
        for (int i = 0; i < n; i++) {
          cout << v[i] << " ";
        }
      } else {
        cout << "NO";
      }
      cout << "\n";
    } else {
      cout << "NO"
           << "\n";
    }
  }
}

```