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: /*package whatever //do not write package name here */
  
  import java.io.*;
  import java.util.*;
  
  public class GFG {
      
  	public static void main (String[] args)throws IOException {
  	     BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
   
          int t=Integer.parseInt(br.readLine());
         
          while(t-->0)
          {
             
              int n = Integer.parseInt(br.readLine());
              String str[] = br.readLine().split(" ");
              long a[] = new long[n];
              long sum = 0;
              for(int i=0;i<n;i++)
              {
                  a[i]=Long.parseLong(str[i]);
                  sum+=a[i];
              }
              
              long k= ((long)n*n+n)/2;
              boolean flag = true;
              
              if(sum%k==0)
              {
                  long tsum=0;
                  long id = sum/k;
                  long b[] = new long[n];
                  if((id-a[0]+a[n-1])%n!=0  && (id-a[0]+a[n-1])>0)
                  flag = false;
                  b[0]=Math.abs((id-a[0]+a[n-1])/n);
                  for(int i=1;i<n;i++)
                  {
                      if((id-a[i]+a[i-1])%n!=0 && (id-a[i]+a[i-1])>0)
                      flag = false;
                     
                      b[i]=Math.abs((id-a[i]+a[i-1])/n);
                      
                  }
                  
                  for(int i=0;i<n;i++)
                  {
                      if(b[i]==0)
                      flag=false;
                       tsum=tsum+b[i];
                  }
                  
                  if(tsum!=id)
                  flag=false;
                   
                  
                 
                  
                  if(flag)
                  {
                      System.out.println("YES");
                      for(int i=0;i<n;i++)
                      System.out.print(b[i]+" ");
                      
                      System.out.println();
                  }
                  else
                  {
                      System.out.println("NO");
                  }
                  
              }
              else
              {
                  System.out.println("NO");
              }
            
          }
  	}
  }