#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << x; }
void __print(const char *x) { cerr << x; }
void __print(const string &x) { cerr << x; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ", ";
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) {
    cerr << (f++ ? ", " : "");
    __print(i);
  }
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) {
    cerr << ", ";
  }
  _print(v...);
}
const long long int MAX_N = 1e5 + 5;
const long long int MOD = 1e9 + 7;
const long long int INF = 1e9;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  long long int sum = 0;
  for (long long int &i : a) {
    cin >> i;
    sum += i;
  }
  long long int val = n * (n + 1) / 2;
  if (sum % val) {
    cout << "NO"
         << "\n";
    ;
  } else {
    sum /= val;
    vector<long long int> ans(n);
    for (long long int i = 0; i < n; i++) {
      long long int dif = a[i] - a[(n - 1 + i) % n];
      dif = sum - dif;
      if (dif <= 0 || (dif % n)) {
        cout << "NO"
             << "\n";
        ;
        return;
      }
      dif /= n;
      ans[i] = dif;
    }
    cout << "YES"
         << "\n";
    ;
    for (long long int i = 0; i < n; i++) cout << ans[i] << ' ';
    cout << "\n";
    ;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(10);
  cout << fixed;
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
}

### 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;
void __print(int x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << x; }
void __print(const char *x) { cerr << x; }
void __print(const string &x) { cerr << x; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ", ";
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) {
    cerr << (f++ ? ", " : "");
    __print(i);
  }
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) {
    cerr << ", ";
  }
  _print(v...);
}
const long long int MAX_N = 1e5 + 5;
const long long int MOD = 1e9 + 7;
const long long int INF = 1e9;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  long long int sum = 0;
  for (long long int &i : a) {
    cin >> i;
    sum += i;
  }
  long long int val = n * (n + 1) / 2;
  if (sum % val) {
    cout << "NO"
         << "\n";
    ;
  } else {
    sum /= val;
    vector<long long int> ans(n);
    for (long long int i = 0; i < n; i++) {
      long long int dif = a[i] - a[(n - 1 + i) % n];
      dif = sum - dif;
      if (dif <= 0 || (dif % n)) {
        cout << "NO"
             << "\n";
        ;
        return;
      }
      dif /= n;
      ans[i] = dif;
    }
    cout << "YES"
         << "\n";
    ;
    for (long long int i = 0; i < n; i++) cout << ans[i] << ' ';
    cout << "\n";
    ;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(10);
  cout << fixed;
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
}

```