#include <bits/stdc++.h>
using namespace std;
long long int MOD = 1000000007;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
unsigned long long power(unsigned long long x, long long int y,
                         long long int p) {
  unsigned long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
unsigned long long modInverse(unsigned long long n, long long int p) {
  return power(n, p - 2, p);
}
unsigned long long nCrModPFermat(unsigned long long n, long long int r,
                                 long long int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  unsigned long long fac[n + 1];
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int find_set(long long int i, vector<long long int> &parent) {
  if (parent[i] == -1) {
    return i;
  } else {
    parent[i] = find_set(parent[i], parent);
    return parent[i];
  }
}
void union_set(long long int i, long long int j, vector<long long int> &parent,
               vector<long long int> &rank) {
  long long int s1 = find_set(i, parent);
  long long int s2 = find_set(j, parent);
  if (s1 != s2) {
    if (rank[s1] > rank[s2]) {
      parent[s2] = s1;
      rank[s1] += rank[s2];
    } else {
      parent[s1] = s2;
      rank[s2] += rank[s1];
    }
  } else {
    return;
  }
}
void solve() {
  long long int n, i, j;
  cin >> n;
  long long int arr[n];
  long long int sum = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  if ((2 * sum) % (n * (n + 1)) != 0) {
    cout << "NO" << endl;
    return;
  }
  long long int val = (2 * sum) / (n * (n + 1));
  if (val < n) {
    cout << "NO" << endl;
    return;
  }
  vector<long long int> a(n);
  for (i = 0; i < n; i++) {
    long long int val1 = arr[i];
    long long int val2 = arr[(i - 1 + n) % (n)];
    if ((val - val1 + val2) <= 0 || (val - val1 + val2) % n != 0) {
      cout << "NO" << endl;
      return;
    }
    a[i] = (val - val1 + val2) / n;
  }
  cout << "YES" << endl;
  for (i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) solve();
}

### 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>
using namespace std;
long long int MOD = 1000000007;
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
unsigned long long power(unsigned long long x, long long int y,
                         long long int p) {
  unsigned long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
unsigned long long modInverse(unsigned long long n, long long int p) {
  return power(n, p - 2, p);
}
unsigned long long nCrModPFermat(unsigned long long n, long long int r,
                                 long long int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  unsigned long long fac[n + 1];
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
long long int find_set(long long int i, vector<long long int> &parent) {
  if (parent[i] == -1) {
    return i;
  } else {
    parent[i] = find_set(parent[i], parent);
    return parent[i];
  }
}
void union_set(long long int i, long long int j, vector<long long int> &parent,
               vector<long long int> &rank) {
  long long int s1 = find_set(i, parent);
  long long int s2 = find_set(j, parent);
  if (s1 != s2) {
    if (rank[s1] > rank[s2]) {
      parent[s2] = s1;
      rank[s1] += rank[s2];
    } else {
      parent[s1] = s2;
      rank[s2] += rank[s1];
    }
  } else {
    return;
  }
}
void solve() {
  long long int n, i, j;
  cin >> n;
  long long int arr[n];
  long long int sum = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  if ((2 * sum) % (n * (n + 1)) != 0) {
    cout << "NO" << endl;
    return;
  }
  long long int val = (2 * sum) / (n * (n + 1));
  if (val < n) {
    cout << "NO" << endl;
    return;
  }
  vector<long long int> a(n);
  for (i = 0; i < n; i++) {
    long long int val1 = arr[i];
    long long int val2 = arr[(i - 1 + n) % (n)];
    if ((val - val1 + val2) <= 0 || (val - val1 + val2) % n != 0) {
      cout << "NO" << endl;
      return;
    }
    a[i] = (val - val1 + val2) / n;
  }
  cout << "YES" << endl;
  for (i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) solve();
}

```