#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cout << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cout << ' ' << H;
  dbg_out(T...);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long sum = accumulate((a).begin(), (a).end(), 0ll);
    long long tot = (n * (n + 1)) / 2;
    if (sum % tot || sum / tot < n) {
      cout << "NO\n";
      continue;
    }
    sum /= tot;
    vector<long long> ans(n);
    long long bad = 0;
    for (long long i = 0; i < n; i++) {
      long long diff = a[(i + 1) % n] - a[i];
      if ((sum - diff) % n) bad = 1;
      ans[(i + 1) % n] = (sum - diff) / n;
    };
    long long s = 0;
    for (auto x : ans) bad |= (x <= 0), s += x;
    if (bad || sum != s)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (auto x : ans) cout << x << " ";
      cout << "\n";
    }
  }
}

### Prompt

Generate a Cpp solution 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;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cout << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cout << ' ' << H;
  dbg_out(T...);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long sum = accumulate((a).begin(), (a).end(), 0ll);
    long long tot = (n * (n + 1)) / 2;
    if (sum % tot || sum / tot < n) {
      cout << "NO\n";
      continue;
    }
    sum /= tot;
    vector<long long> ans(n);
    long long bad = 0;
    for (long long i = 0; i < n; i++) {
      long long diff = a[(i + 1) % n] - a[i];
      if ((sum - diff) % n) bad = 1;
      ans[(i + 1) % n] = (sum - diff) / n;
    };
    long long s = 0;
    for (auto x : ans) bad |= (x <= 0), s += x;
    if (bad || sum != s)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (auto x : ans) cout << x << " ";
      cout << "\n";
    }
  }
}

```