#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
long long A[N], sum[N], dif[N], dif1[N], ans[N];
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
    if (i > 1) sum[i] = A[i] - A[i - 1];
  }
  sum[1] = A[1] - A[n];
  for (int i = 2; i <= n; ++i) {
    dif[i] = sum[i - 1] - sum[i];
    if (dif[i] % n) {
      cout << "NO\n";
      return;
    }
    dif[i] /= n;
  }
  for (int i = 2; i <= n; ++i) {
    dif1[i] = dif1[i - 1] + dif[i];
  }
  long long k = 1, b = 0;
  for (int i = 2; i <= n; ++i) {
    k += (n - i + 2);
    b += dif1[i] * (n - i + 2);
  }
  if ((A[1] - b) % k) {
    cout << "NO\n";
    return;
  }
  ans[1] = (A[1] - b) / k;
  for (int i = 1; i <= n; ++i) {
    ans[i] = ans[1] + dif1[i];
    if (ans[i] < 1) {
      cout << "NO\n";
      return;
    }
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) cout << ans[1] + dif1[i] << ' ';
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

### 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 int N = 2e5 + 7;
long long A[N], sum[N], dif[N], dif1[N], ans[N];
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
    if (i > 1) sum[i] = A[i] - A[i - 1];
  }
  sum[1] = A[1] - A[n];
  for (int i = 2; i <= n; ++i) {
    dif[i] = sum[i - 1] - sum[i];
    if (dif[i] % n) {
      cout << "NO\n";
      return;
    }
    dif[i] /= n;
  }
  for (int i = 2; i <= n; ++i) {
    dif1[i] = dif1[i - 1] + dif[i];
  }
  long long k = 1, b = 0;
  for (int i = 2; i <= n; ++i) {
    k += (n - i + 2);
    b += dif1[i] * (n - i + 2);
  }
  if ((A[1] - b) % k) {
    cout << "NO\n";
    return;
  }
  ans[1] = (A[1] - b) / k;
  for (int i = 1; i <= n; ++i) {
    ans[i] = ans[1] + dif1[i];
    if (ans[i] < 1) {
      cout << "NO\n";
      return;
    }
  }
  cout << "YES\n";
  for (int i = 1; i <= n; ++i) cout << ans[1] + dif1[i] << ' ';
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}

```