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. 
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import java.util.*;

public class Main
{
	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];
		    
		    long sum = 0;
		    for(int i =0;i<n;i++)
		    {
		        b[i] = sc.nextInt();
		        sum += b[i];
		    }
		    
		    
		    if(sum%(n*(n+1)/2) != 0)
		    {
		        System.out.println("NO");
		    }
		    else
		    {
		        long a[] = new long[n];
		        
		        sum = sum/(n*(n+1)/2);
		        
		        int flag = 1;
    		        for(int i = 0;i<n;i++)
    		        {
    		            if(i == 0)
    		            {
    		                if((sum + b[n-1] - b[i])%n == 0 && (sum + b[n-1] - b[i])/n > 0)
    		                    a[i] = (sum + b[n-1] - b[i])/n;
    		                else
    		                {
    		                    //System.out.println(a[i]);
    		                    flag = 0;
    		                    break;
    		                }
    		            }
    		            else
    		            {
    		                if((sum + b[i-1] - b[i])%n == 0 && (sum + b[i-1] - b[i])/n > 0)
    		                    a[i] = (sum + b[i-1] - b[i])/n;
    		                else
    		                {
    		                    //System.out.println(a[i]);
    		                    
    		                    flag = 0;
    		                    break;
    		                }
    		                
    		            }
    		        }
		        
		        if(flag == 0)
		            System.out.println("NO");
		       else{
    		        System.out.println("YES");
    		        for(int i = 0;i<n;i++)
    		        {
    		            System.out.print(a[i]+" ");
    		        }
    		        System.out.println();
		       }
		    }
		    
		    
		}
		
	}
}
