#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long N = 2e3 + 100;
const long long MOD = 1e9 + 7;
const long long inf = 2e9;
const long double eps = 1e-12;
const long double EPS = 1e-10;
long long rx[] = {1, -1, 0, 0};
long long ry[] = {0, 0, 1, -1};
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    long long sum = 0;
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
      sum += a[i];
    }
    long long k = n * (n + 1) / 2;
    if (sum % k) {
      cout << "NO" << '\n';
      continue;
    }
    long long s = sum / k;
    vector<long long> v(n);
    bool ok = 1;
    for (long long i = 0; i < n && ok; ++i) {
      long long j = i - 1;
      if (j < 0) j = n - 1;
      long long x = a[i] - a[j] - s;
      x = -x;
      if (x % n || x / n <= 0)
        ok = 0;
      else
        v[i] = x / n;
    }
    if (!ok) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES\n";
    for (auto to : v) cout << to << ' ';
    cout << '\n';
  }
  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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long N = 2e3 + 100;
const long long MOD = 1e9 + 7;
const long long inf = 2e9;
const long double eps = 1e-12;
const long double EPS = 1e-10;
long long rx[] = {1, -1, 0, 0};
long long ry[] = {0, 0, 1, -1};
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    long long sum = 0;
    for (long long i = 0; i < n; ++i) {
      cin >> a[i];
      sum += a[i];
    }
    long long k = n * (n + 1) / 2;
    if (sum % k) {
      cout << "NO" << '\n';
      continue;
    }
    long long s = sum / k;
    vector<long long> v(n);
    bool ok = 1;
    for (long long i = 0; i < n && ok; ++i) {
      long long j = i - 1;
      if (j < 0) j = n - 1;
      long long x = a[i] - a[j] - s;
      x = -x;
      if (x % n || x / n <= 0)
        ok = 0;
      else
        v[i] = x / n;
    }
    if (!ok) {
      cout << "NO" << '\n';
      continue;
    }
    cout << "YES\n";
    for (auto to : v) cout << to << ' ';
    cout << '\n';
  }
  return 0;
}

```