#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y);
long long int gcd(long long int a, long long int b);
vector<long long int> ft = {1};
long long int modInv(long long int i) { return power(i, 1000000007 - 2); }
long long int ncr(long long int n, long long int r) {
  return (n >= r ? (ft[n] * modInv(ft[r])) % 1000000007 * modInv(ft[n - r]) %
                       1000000007
                 : 0);
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x %= 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    long long int s = 0;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      s += a[i];
    }
    long long int val = (n * (n + 1)) / 2;
    if (s % val != 0) {
      cout << "NO\n";
      continue;
    }
    val = s / val;
    long long int f = 0;
    vector<long long int> ans;
    for (long long int i = 0; i < n; i++) {
      long long int tmp = val - a[i] + a[(i - 1 + n) % n];
      if (tmp < 1 or tmp % n) {
        f = 1;
        break;
      } else {
        ans.push_back(tmp / n);
      }
    }
    if (f)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (auto i : ans) cout << i << " ";
      cout << "\n";
    }
  }
}

### Prompt

Please create a solution in CPP 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 int power(long long int x, long long int y);
long long int gcd(long long int a, long long int b);
vector<long long int> ft = {1};
long long int modInv(long long int i) { return power(i, 1000000007 - 2); }
long long int ncr(long long int n, long long int r) {
  return (n >= r ? (ft[n] * modInv(ft[r])) % 1000000007 * modInv(ft[n - r]) %
                       1000000007
                 : 0);
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x %= 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res % 1000000007;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    long long int s = 0;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      s += a[i];
    }
    long long int val = (n * (n + 1)) / 2;
    if (s % val != 0) {
      cout << "NO\n";
      continue;
    }
    val = s / val;
    long long int f = 0;
    vector<long long int> ans;
    for (long long int i = 0; i < n; i++) {
      long long int tmp = val - a[i] + a[(i - 1 + n) % n];
      if (tmp < 1 or tmp % n) {
        f = 1;
        break;
      } else {
        ans.push_back(tmp / n);
      }
    }
    if (f)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (auto i : ans) cout << i << " ";
      cout << "\n";
    }
  }
}

```