#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
int biExp(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int res = 1;
  while (n) {
    if (n & 1) res *= a, res %= M;
    a *= a, a %= M;
    n >>= 1;
  }
  return res;
}
int modMul(long long int a, long long int n) {
  if (n == 0) return 0;
  long long int res = 0;
  while (n) {
    if (n & 1) res += a, res %= M;
    a += a, a %= M;
    n >>= 1;
  }
  return res;
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " || ";
  __f(comma + 1, args...);
}
void prakashitBhavah() {
  int n;
  cin >> n;
  long long int total_b = 0, total_a = 0;
  vector<int> b(n), a(n);
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    total_b += b[i];
  }
  int x = n * (n + 1) / 2;
  if (total_b % x != 0) {
    cout << "NO\n";
    return;
  }
  total_a = total_b / x;
  bool flag = true;
  if (total_a < n) {
    cout << "NO\n";
    return;
  }
  long long int temp_a = 0;
  for (int i = 1; i < n; i++) {
    int diff = total_a - (b[i] - b[i - 1]);
    if (diff % n != 0) flag = false;
    a[i] = (diff) / n;
    temp_a += a[i];
  }
  a[0] = total_a - temp_a;
  for (int x : a) {
    if (x <= 0) flag = false;
  }
  if (flag) {
    cout << "YES\n";
    for (int x : a) cout << x << " ";
    cout << "\n";
  } else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) prakashitBhavah();
  return 0;
}

### 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>
using namespace std;
const int M = 1e9 + 7;
int biExp(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int res = 1;
  while (n) {
    if (n & 1) res *= a, res %= M;
    a *= a, a %= M;
    n >>= 1;
  }
  return res;
}
int modMul(long long int a, long long int n) {
  if (n == 0) return 0;
  long long int res = 0;
  while (n) {
    if (n & 1) res += a, res %= M;
    a += a, a %= M;
    n >>= 1;
  }
  return res;
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " || ";
  __f(comma + 1, args...);
}
void prakashitBhavah() {
  int n;
  cin >> n;
  long long int total_b = 0, total_a = 0;
  vector<int> b(n), a(n);
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    total_b += b[i];
  }
  int x = n * (n + 1) / 2;
  if (total_b % x != 0) {
    cout << "NO\n";
    return;
  }
  total_a = total_b / x;
  bool flag = true;
  if (total_a < n) {
    cout << "NO\n";
    return;
  }
  long long int temp_a = 0;
  for (int i = 1; i < n; i++) {
    int diff = total_a - (b[i] - b[i - 1]);
    if (diff % n != 0) flag = false;
    a[i] = (diff) / n;
    temp_a += a[i];
  }
  a[0] = total_a - temp_a;
  for (int x : a) {
    if (x <= 0) flag = false;
  }
  if (flag) {
    cout << "YES\n";
    for (int x : a) cout << x << " ";
    cout << "\n";
  } else
    cout << "NO\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) prakashitBhavah();
  return 0;
}

```