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.lang.Math;
import java.io.* ;


public class Account {
	
	public static class Pair<Object1 ,Object2> {
	    Object1 key ;
	    Object2 val ;
	    
	    Pair(Object1 key ,Object2 val) {
	        this.key = key ;
	        this.val = val ;
	    }
	    
	    Pair() {}
	    
	}
	
	public static void main (String[] args) {
		
//		Scanner sc = new Scanner(System.in) ;
		FastReader sc = new FastReader() ;
//		StringBuilder res = new StringBuilder() ;
		int t= 1 ;  
		t = sc.nextInt() ;
		while(t-- > 0) {
			int n = sc.nextInt() ;
			long[] arr = new long[n] ;
			long sum= 0 ;
			for(int i=0 ;i<n ;i++) {
				arr[i] = sc.nextLong() ;
				sum += arr[i] ;
			}
			sum = (sum*2) / (n*(n+1)) ;
			long[] ans = new long[n] ;
			boolean flag = true ;
			int prv = n-1 ;
			for(int i=0 ;i<n ;i++) {
				long val = arr[prv] + sum - arr[i] ;
				if(val%n != 0  ||  val <= 0) {
					flag = false ;
					break ;
				}else {
					ans[i] = val / n ; ;
					prv = i ;
				}
			}if(flag == true) {
				System.out.println("YES");
				for(int i=0 ;i<n ;i++) {
					System.out.print(ans[i]+" ");
				}
				System.out.println();
			}else {
				System.out.println("NO");
			}
  		}
	}
	
	
	
	
	/*
	 *   1--> a   3b   2c
	 *   2--> 2a  b    3c
	 *   3--> 3a  2b   c
	 */
		
	
	public static long gcd(long a, long b)
	{
	    if (b == 0)
	        return a;
	    return gcd(b, a % b);
	     
	}
	
	public static int xor(int n)
    {
        if (n % 4 == 0)
            return n;
      
        if (n % 4 == 1)
            return 1;
      
        if (n % 4 == 2)
            return n + 1;
      
        return 0;
    }

	
	public static class FastReader {
        
		BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt()
        {
            return Integer.parseInt(next());
        }
 
        long nextLong()
        {
            return Long.parseLong(next());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
 
        int [] readintarray(int n) {
            int res [] = new int [n];
            for(int i = 0; i<n; i++)res[i] = nextInt();
            return res;
        }
        long [] readlongarray(int n) {
            long res [] = new long [n];
            for(int i = 0; i<n; i++)res[i] = nextLong();
            return res;
        }
    }	
	
}