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. 
import java.util.Scanner;

public class eqN {
    static void printArr(int[] a){
        for(int i=0; i<a.length; i++){
            System.out.print(a[i]+" ");
        }
        System.out.println();
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t= sc.nextInt();
        while(t-->0){
            int n= sc.nextInt();
            int[] b= new int[n];
            int sum = 0;
            for(int i=0; i<n; i++) {
                b[i]= sc.nextInt();
                sum+=b[i];
            }
            if(n==1){
                System.out.println("YES");
                System.out.println(b[0]);
                continue;
            }
            int[] ar = new int[n-1];
            for(int i=0; i<n-1; i++) {
                ar[i]=b[i+1]-b[i];
            }
            int tmp=0;
            int tmp2= n*(n+1)/2;
            int[] ans = new int[n];
            //System.out.println();
            if(sum%tmp2==0&&sum>tmp2){
                sum/=tmp2;
                System.out.println("YES");
                for(int i=0; i<n-1; i++) {
                    ans[i+1]= (sum-ar[i])/n;
                    tmp+=ans[i+1];
                }
                ans[0]=sum-tmp;
                printArr(ans);
            }else{
                System.out.println("NO");
            }

        }
    }
}
