#include <bits/stdc++.h>
using namespace std;
long long n;
long long a[400010];
long long sum = 0;
void read() {
  cin >> n;
  sum = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
}
void solve() {
  vector<long long> ans;
  long long hs = n * (n + 1) / 2;
  if (sum % hs != 0) {
    cout << "NO" << '\n';
    return;
  }
  long long total = sum / hs;
  if (total < n) {
    cout << "NO" << '\n';
    return;
  }
  a[0] = a[n];
  for (int i = 1; i <= n; i++) {
    long long tmp = a[i] - a[i - 1];
    long long x = total - tmp;
    if (x <= 0) {
      cout << "NO" << '\n';
      return;
    }
    if (x % n != 0) {
      cout << "NO" << '\n';
      return;
    }
    x = x / n;
    ans.push_back(x);
  }
  long long curSum = 0;
  for (auto i : ans) curSum += i;
  if (curSum != total) {
    cout << "NO" << '\n';
    return;
  }
  cout << "YES" << '\n';
  for (auto i : ans) cout << i << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    read();
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long n;
long long a[400010];
long long sum = 0;
void read() {
  cin >> n;
  sum = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
}
void solve() {
  vector<long long> ans;
  long long hs = n * (n + 1) / 2;
  if (sum % hs != 0) {
    cout << "NO" << '\n';
    return;
  }
  long long total = sum / hs;
  if (total < n) {
    cout << "NO" << '\n';
    return;
  }
  a[0] = a[n];
  for (int i = 1; i <= n; i++) {
    long long tmp = a[i] - a[i - 1];
    long long x = total - tmp;
    if (x <= 0) {
      cout << "NO" << '\n';
      return;
    }
    if (x % n != 0) {
      cout << "NO" << '\n';
      return;
    }
    x = x / n;
    ans.push_back(x);
  }
  long long curSum = 0;
  for (auto i : ans) curSum += i;
  if (curSum != total) {
    cout << "NO" << '\n';
    return;
  }
  cout << "YES" << '\n';
  for (auto i : ans) cout << i << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    read();
    solve();
  }
  return 0;
}

```