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.*;
import java.io.*;

////***************************************************************************
 /* public class E_Gardener_and_Tree implements Runnable{

       public static void main(String[] args) throws Exception {
        new Thread(null, new E_Gardener_and_Tree(), "E_Gardener_and_Tree", 1<<28).start();
       }
    public void run(){
         WRITE YOUR CODE HERE!!!!
         JUST WRITE EVERYTHING HERE WHICH YOU WRITE IN MAIN!!!
       }

  }
*/
/////**************************************************************************


public class E_Singers_Tour{
    public static void main(String[] args) {
        FastScanner s= new FastScanner();
        //PrintWriter out=new PrintWriter(System.out);
        //end of program
        //out.println(answer);
        //out.close();
        StringBuilder res = new StringBuilder();
        int t=s.nextInt();
        int p=0;
        while(p<t){
        int n=s.nextInt();
        long array[]= new long[n];
        long sum=0;
        for(int i=0;i<n;i++){
            array[i]=s.nextLong();
            sum+=array[i];
        }
        long yo=n;
        long a1=2;
        long hh=((yo)*(yo+1))/a1;
        
        if(sum%hh!=0){
            res.append("NO \n");
        }
        else{
            long well=sum/hh;
            long ans[]= new long[n];
            int check=0;
            for(int i=1;i<n;i++){
                long yo1=array[i-1]-array[i]+well;
                if(yo1<=0){
                    check=1;
                    break;
                }
                else{
                    if(yo1%n!=0){
                        check=1;
                        break;
                    }
                    else{
                        long kk=yo1/n;
                        ans[i]=kk;
                    }
                }

            }
            long q=array[n-1]-array[0]+well;
            if(q<=0){
                check=1;
            }
            if(q%n!=0){
                check=1;
            }
            else{
                long kk=q/n;
                ans[0]=kk;
            }
            if(check==0){
                res.append("YES \n");
                for(int i=0;i<n;i++){
                    res.append(ans[i]+" ");
                }
                res.append(" \n");
            }
            else{
                res.append("NO \n");
            }

        }
        p++;
    }
    System.out.println(res);
    }
    static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
 
		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}
 
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(nextToken());
		}
 
		long nextLong() {
			return Long.parseLong(nextToken());
		}
 
		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
	}


    static long modpower(long x, long y, long p)
     {
      long res = 1; // Initialize result
   
      x = x % p; // Update x if it is more than or
      // equal to p
   
      if (x == 0)
        return 0; // In case x is divisible by p;
   
      while (y > 0)
      {
   
        // If y is odd, multiply x with result
        if ((y & 1) != 0)
          res = (res * x) % p;
   
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
      }
      return res;
    }

  
   // SIMPLE POWER FUNCTION=>


      static long power(long x, long y)
        {
            long res = 1;     // Initialize result
        
            while (y > 0)
            {
            
                // If y is odd, multiply x with result
                if ((y & 1) != 0)
                    res = res * x;
        
                // y must be even now
                y = y >> 1; // y = y/2
                x = x * x;  // Change x to x^2
            }
            return res;
        }





}