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.io.*;
import java.util.*;
public class solution {
	static long cr[][]=new long[1001][1001];
   static long mod=1000000007;
public static void main(String[] args)  {
	FScanner sc = new FScanner();
	//Arrays.fill(prime, true);
	//sieve();
	
	//ncr();
	int t=sc.nextInt();
	 StringBuilder sb = new StringBuilder();
	while(t-->0) {	
	int n=sc.nextInt();
	int a[]=new int[n];
	long sum=0;
	for(int i=0;i<n;i++)
		{
		a[i]=sc.nextInt();
		sum=sum+a[i];
		}
	long val_sum=(n*(n+1))/2;
	long ans[]=new long[n+1];
	if(sum%val_sum==0)
	{
		sum=sum/val_sum;
		for(int i=0;i<n;i++)
		{
			long dur= sum -(a[(i+1)%n] - a[i]);
			if(dur%n!=0)
				dur=0;
			else
				dur=dur/n;
			ans[i+1]=dur;
		}
		ans[0]=ans[n];
		boolean flag=true;
		for(int i=0;i<n;i++)
		{
			if(ans[i]==0)
			{
				flag=false;
				sb.append("NO");
				break;
			}
		}
		
		if(flag)
		{  sb.append("YES"+"\n");
			for(int i=0;i<n;i++)
			{
				sb.append(ans[i]+" ");
			}
		}
		
		
	}
	else
	{
		sb.append("NO");
	}
	
    sb.append("\n");
			
		}
		
	System.out.println(sb.toString());	
}




public static long gcd(long a,long b)
{   
	
     return b==0 ? a:gcd(b,a%b);	
	
}

        
 /* public static void sieve()
{   prime[0]=prime[1]=false;
int n=1000000;
	for(int p = 2; p*p <=n; p++)
    {
        if(prime[p] == true)
        {
          
            for(int i = p*p; i <= n; i += p)
                prime[i] = false;
        }
    }
   */

public static void ncr()
{
	cr[0][0]=1;
	for(int i=1;i<=1000;i++)
	{
		cr[i][0]=1;
		for(int j=1;j<i;j++)
		{
			cr[i][j]=(cr[i-1][j-1]+cr[i-1][j])%mod;
		}
		cr[i][i]=1;
	}
}

}
 
  class pair {
	int val;int ind;
	 pair(int val,int ind)
	{
		this.val=val;
		this.ind=ind;
	}
	
	
}  
 
 class FScanner {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer sb = new StringTokenizer("");
 
	String next(){
		while(!sb.hasMoreTokens()){
			try{
				sb = new StringTokenizer(br.readLine());
			} catch(IOException e){ }
		}
		return sb.nextToken();
	}
	String nextLine(){
		try{ return br.readLine(); } 
		catch(IOException e) { } return "";
	}
 
	int nextInt(){
		return Integer.parseInt(next());
	}
 
	long nextLong() {
		return Long.parseLong(next());
	}
 
	int[] readArray(int n) {
		int a[] = new int[n];
		for(int i=0;i<n;i++) a[i] = nextInt();
		return a;
	}
 
	float nextFloat(){
	return Float.parseFloat(next());
	}
	
	double nextDouble(){
	return Double.parseDouble(next());
	}
}