#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int limit = 5e5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(long long x, char c = '\n') {
  write(x);
  putchar(c);
}
const int mod = 1e9 + 7;
const int M = 500 + 5;
const int S = (1 << 13) + 5;
const int N = 2e5 + 5;
int n, m, _, k;
long long a[N];
long long b[N], d[N];
signed main() {
  int T;
  while (cin >> T)
    while (T--) {
      scanf("%d", &n);
      for (int i = (1); i <= (n); i++) scanf("%lld", &b[i]);
      long long sum = 0;
      for (int i = (1); i <= (n); i++) sum += b[i];
      int ok = 1;
      int cnt = n * (n + 1) / 2;
      if (sum % cnt)
        ok = 0;
      else {
        sum = sum / cnt;
        for (int i = 1; i <= n && ok; i++) {
          int p = i - 1;
          if (p == 0) p = n;
          d[i] = b[i] - b[p] - sum;
          if (d[i] % n || d[i] >= 0) {
            ok = 0;
          } else {
            d[i] /= n;
            d[i] = -d[i];
          }
        }
      }
      if (ok) {
        puts("YES");
        for (int i = (1); i <= (n); i++) printf("%lld ", d[i]);
        puts("");
      } else {
        puts("NO");
      }
    }
  return 0;
}

### 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;
const double pi = acos(-1.0);
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const int limit = 5e5;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(long long x, char c = '\n') {
  write(x);
  putchar(c);
}
const int mod = 1e9 + 7;
const int M = 500 + 5;
const int S = (1 << 13) + 5;
const int N = 2e5 + 5;
int n, m, _, k;
long long a[N];
long long b[N], d[N];
signed main() {
  int T;
  while (cin >> T)
    while (T--) {
      scanf("%d", &n);
      for (int i = (1); i <= (n); i++) scanf("%lld", &b[i]);
      long long sum = 0;
      for (int i = (1); i <= (n); i++) sum += b[i];
      int ok = 1;
      int cnt = n * (n + 1) / 2;
      if (sum % cnt)
        ok = 0;
      else {
        sum = sum / cnt;
        for (int i = 1; i <= n && ok; i++) {
          int p = i - 1;
          if (p == 0) p = n;
          d[i] = b[i] - b[p] - sum;
          if (d[i] % n || d[i] >= 0) {
            ok = 0;
          } else {
            d[i] /= n;
            d[i] = -d[i];
          }
        }
      }
      if (ok) {
        puts("YES");
        for (int i = (1); i <= (n); i++) printf("%lld ", d[i]);
        puts("");
      } else {
        puts("NO");
      }
    }
  return 0;
}

```