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;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less_equal<int>, rb_tree_tag,tree_order_statistics_node_update>
#define int long long
#define mod 1000000007
#define PI 3.1415926535
#define maxn 100005
#define rall(x) x.rbegin(), x.rend()
#define all(x) x.begin(),x.end()
#define speedUp ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
 
void _SSGJ_()
{   
    int n;
    cin>>n;

    vector <int> v(n);
    int s=0;
    for(int i=0;i<n;i++){
        cin>>v[i];
        s += v[i];
    }

    int val = n*(n + 1)/2;
    if(s%val){
        cout<<"NO\n";
        return;
    }

    s /= val;
    vector <int> ans(n,0);
    for(int i=0;i<n - 1;i++){
        int d = v[i + 1] - v[i];
        if((s - d)%n || s - d <= 0){
            cout<<"NO\n";
            return;
        }

        ans[i + 1] = (s - d)/n;
    }

    for(auto i: ans){
        s -= i;
    }

    ans[0] = s;
    cout<<"YES\n";
    for(auto i: ans){
        cout<<i<<" ";
    }
    cout<<"\n";

}
 
signed main()
{
    speedUp
    ios_base::sync_with_stdio(false);cin.tie(NULL);
   
   #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("error.txt", "w", stderr);
    freopen("output.txt", "w", stdout);
   #endif
   
    int t;
    cin>>t;
    while(t--)
    _SSGJ_();
   
    cerr<<"time taken : "<<(float)clock()/CLOCKS_PER_SEC<<" secs"<<"\n";
    return 0;
}