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.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() ;
  			double[] arr = new double[n] ;
  			double sum= 0 ;
  			for(int i=0 ;i<n ;i++) {
  				arr[i] = sc.nextDouble() ;
  				sum += arr[i] ;
  			}
  			sum = (sum*2) / (n*(n+1)) ;
  			if(sum != (long)(sum)) {
  				System.out.println("NO");
  			}else {
  				long[] ans = new long[n] ;
  				boolean flag = true ;
  				int prv = n-1 ;
  				for(int i=0 ;i<n ;i++) {
  					double val = arr[prv] + sum - arr[i] ;
  					val = val / n ;
  					if(val != (long)(val)  ||  val <= 0) {
  						flag = false ;
  						break ;
  					}else {
  						ans[i] = (long)val ;
  						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;
          }
      }	
  	
  }