#include <bits/stdc++.h>
using namespace std;
const int N = 4e4 + 10;
const int inf = 1e9;
int t;
int n;
int b[N], ans[N];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  long long tot = 0;
  for (int i = 0; i < n; i++) tot += b[i];
  if (tot % (n * (n + 1) / 2) != 0) {
    cout << "NO\n";
    return;
  }
  tot /= n * (n + 1) / 2;
  for (int i = 0; i < n; i++) {
    long long tmp = b[(i - 1 + n) % n] - b[i] + tot;
    if (tmp <= 0 || tmp % n != 0 || tmp / n > inf) {
      cout << "NO\n";
      return;
    }
    ans[i] = tmp / n;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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 int N = 4e4 + 10;
const int inf = 1e9;
int t;
int n;
int b[N], ans[N];
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  long long tot = 0;
  for (int i = 0; i < n; i++) tot += b[i];
  if (tot % (n * (n + 1) / 2) != 0) {
    cout << "NO\n";
    return;
  }
  tot /= n * (n + 1) / 2;
  for (int i = 0; i < n; i++) {
    long long tmp = b[(i - 1 + n) % n] - b[i] + tot;
    if (tmp <= 0 || tmp % n != 0 || tmp / n > inf) {
      cout << "NO\n";
      return;
    }
    ans[i] = tmp / n;
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```