#include <bits/stdc++.h>
using namespace std;
long long MAX = 1e9 + 7;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
long long powM(long long x, long long y, long long m) {
  long long ans = 1, r = 1;
  x = x % m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans = ans * x;
      ans = ans % m;
    }
    r = r << 1;
    x = x * x;
    x = x % m;
  }
  ans = ans % m;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long T = 0; T < t; T++) {
    long long n;
    cin >> n;
    long long b[n];
    for (long long i = 0; i < n; i++) cin >> b[i];
    long long a[n];
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      sum += b[i];
    }
    bool imp = 0;
    if ((sum) % (n * (n + 1) / 2) == 0)
      sum = (2 * sum) / (n * (n + 1));
    else {
      imp = 1;
    }
    if (!imp) {
      for (long long i = 0; i < n; i++) {
        if ((b[(i - 1 + n) % n] - b[i % n] + sum) % n == 0)
          a[i] = (b[(i - 1 + n) % n] - b[i % n] + sum) / n;
        else {
          imp = 1;
          break;
        }
        if (a[i] < 1 || a[i] > 1000000000) {
          imp = 1;
          break;
        }
      }
    }
    if (imp) {
      cout << "NO" << endl;
      goto done;
    } else {
      cout << "YES" << endl;
    }
    for (long long i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  done:;
  }
  return (0);
}

### Prompt

Please formulate a cpp solution to 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;
long long MAX = 1e9 + 7;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
long long powM(long long x, long long y, long long m) {
  long long ans = 1, r = 1;
  x = x % m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans = ans * x;
      ans = ans % m;
    }
    r = r << 1;
    x = x * x;
    x = x % m;
  }
  ans = ans % m;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long T = 0; T < t; T++) {
    long long n;
    cin >> n;
    long long b[n];
    for (long long i = 0; i < n; i++) cin >> b[i];
    long long a[n];
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      sum += b[i];
    }
    bool imp = 0;
    if ((sum) % (n * (n + 1) / 2) == 0)
      sum = (2 * sum) / (n * (n + 1));
    else {
      imp = 1;
    }
    if (!imp) {
      for (long long i = 0; i < n; i++) {
        if ((b[(i - 1 + n) % n] - b[i % n] + sum) % n == 0)
          a[i] = (b[(i - 1 + n) % n] - b[i % n] + sum) / n;
        else {
          imp = 1;
          break;
        }
        if (a[i] < 1 || a[i] > 1000000000) {
          imp = 1;
          break;
        }
      }
    }
    if (imp) {
      cout << "NO" << endl;
      goto done;
    } else {
      cout << "YES" << endl;
    }
    for (long long i = 0; i < n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  done:;
  }
  return (0);
}

```