description: 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. 
solution: 
  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]<1)
  			{
  				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());
  	}
  }