#include <bits/stdc++.h>
long double eps = 1e-9;
const long long int maxn = 1e17 + 10;
const long long int mod = 998244353;
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = res * a;
    a = a * a;
    b = b / 2;
  }
  return res;
}
long long int po(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = ((res % mod) * (x % mod)) % mod;
    y = y >> 1;
    x = ((x % mod) * (x % mod)) % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<long long int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<long long int> ans(n);
    int flag = 1;
    ans[0] = 0;
    for (int i = 1; i < n; i++) {
      long long int num = a[i - 1] - a[i] - a[n - 1] + a[0];
      if (num % n != 0) flag = 0;
      ans[i] = num / n;
    }
    long long int no = 1, num = a[0];
    for (int i = n - 1; i >= 1; i--) {
      long long int fac = n + 1 - i;
      no += fac;
      num -= (fac * ans[i]);
    }
    if (num % no != 0 || num < 0) flag = 0;
    ans[0] = num / no;
    for (int i = 1; i < n; i++) {
      ans[i] += ans[0];
      if (ans[i] <= 0) flag = 0;
    }
    if (ans[0] <= 0) flag = 0;
    if (flag) {
      cout << "YES" << '\n';
      for (int i = 0; i < n; i++) cout << ans[i] << " ";
      cout << '\n';
    } else
      cout << "NO" << '\n';
  }
}

### Prompt

Create a solution in cpp for 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>
long double eps = 1e-9;
const long long int maxn = 1e17 + 10;
const long long int mod = 998244353;
using namespace std;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = res * a;
    a = a * a;
    b = b / 2;
  }
  return res;
}
long long int po(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = ((res % mod) * (x % mod)) % mod;
    y = y >> 1;
    x = ((x % mod) * (x % mod)) % mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<long long int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<long long int> ans(n);
    int flag = 1;
    ans[0] = 0;
    for (int i = 1; i < n; i++) {
      long long int num = a[i - 1] - a[i] - a[n - 1] + a[0];
      if (num % n != 0) flag = 0;
      ans[i] = num / n;
    }
    long long int no = 1, num = a[0];
    for (int i = n - 1; i >= 1; i--) {
      long long int fac = n + 1 - i;
      no += fac;
      num -= (fac * ans[i]);
    }
    if (num % no != 0 || num < 0) flag = 0;
    ans[0] = num / no;
    for (int i = 1; i < n; i++) {
      ans[i] += ans[0];
      if (ans[i] <= 0) flag = 0;
    }
    if (ans[0] <= 0) flag = 0;
    if (flag) {
      cout << "YES" << '\n';
      for (int i = 0; i < n; i++) cout << ans[i] << " ";
      cout << '\n';
    } else
      cout << "NO" << '\n';
  }
}

```