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. 
#include <bits/stdc++.h>

using namespace std;

#define int                     long long
#define ll                      long long
#define ld                      long double
#define pb                      push_back
#define mp                      make_pair
#define f                       first
#define s                       second
#define all(x)                  (x).begin(), (x).end()
#define LOL                     cout << '\n';
#define kill(x)                 {cout << x << '\n'; return;}
#define sz(x)                   ((int)(x).size())
#define deb(x)                  cout << "[" << (#x) << " = " << (x) << "]" << '\n'
#define deb2(x, y)              cout << "[" << (#x) << " = " << (x) << "] [" << (#y) << " = " << (y) << "]" << '\n'
#define deb3(x, y, z)           cout << "[" << (#x) << " = " << (x) << "] [" << (#y) << " = " << (y) << "] [" << (#z) << " = " << (z) << "]" << '\n'
#define deb4(x, y, z, p)        cout << "[" << (#x) << " = " << (x) << "] [" << (#y) << " = " << (y) << "] [" << (#z) << " = " << (z) << "] [" << (#p) << " = " << (p) << "]"<< '\n'
#define deb5(x, y, z, p, q)     cout << "[" << (#x) << " = " << (x) << "] [" << (#y) << " = " << (y) << "] [" << (#z) << " = " << (z) << "] [" << (#p) << " = " << (p) << "] [" << (#q) << " = " << (q) << "]"<< '\n'
#define deb6(x, y, z, p, q, r)  cout << "[" << (#x) << " = " << (x) << "] [" << (#y) << " = " << (y) << "] [" << (#z) << " = " << (z) << "] [" << (#p) << " = " << (p) << "] [" << (#q) << " = " << (q) << "] [" << (#r) << " = " << (r) << "]"<< '\n'

const int M     = 1e9 + 7;
const int MOD   = 998244353;
const int INF   = 0x7f7f7f7f'7f7f7f7f;
const int Nmax  = 2e5 + 15;


void solve() {

  int N; cin >> N;
  std::vector<int> v(N);
  int S = 0;
  for(auto &el:v) cin>> el, S += el;

  std::vector<int> ok;

  int e = N * (N + 1) / 2;
  
  for(int i = 0; i < N; i++)
  {
    int j = (((i - 1) % N) + N ) % N;
    int y = S + v[j] * e - v[i] * e;
    // deb(y);
    if(y == 0 || y % ((N * N * (N + 1)) / 2) != 0) kill("NO");
    ok.pb((y / (( N * N * (N + 1)) / 2)) );
  }

  cout << "YES"; LOL
  for(auto &el:ok) cout << el << ' ';LOL



}

signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0); cout.tie(0);
  #ifdef Kaushal_26
    freopen("a.in", "r", stdin); 
    freopen("b.out", "w", stdout);  
    freopen("e.out", "w", stderr);
  #endif
  int _ = 1; cin >> _;
  for(int i = 1 ; i <= _ ; i++){
      // cout << "\nCase #" << i << ": \n";
      solve();
  }
  return 0;
}

// 1.77245