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.io.*;
  
  ////***************************************************************************
   /* public class E_Gardener_and_Tree implements Runnable{
  
         public static void main(String[] args) throws Exception {
          new Thread(null, new E_Gardener_and_Tree(), "E_Gardener_and_Tree", 1<<28).start();
         }
      public void run(){
           WRITE YOUR CODE HERE!!!!
           JUST WRITE EVERYTHING HERE WHICH YOU WRITE IN MAIN!!!
         }
  
    }
  */
  /////**************************************************************************
  
  
  public class E_Singers_Tour{
      public static void main(String[] args) {
          FastScanner s= new FastScanner();
          //PrintWriter out=new PrintWriter(System.out);
          //end of program
          //out.println(answer);
          //out.close();
          StringBuilder res = new StringBuilder();
          int t=s.nextInt();
          int p=0;
          while(p<t){
          int n=s.nextInt();
          long array[]= new long[n];
          long sum=0;
          for(int i=0;i<n;i++){
              array[i]=s.nextLong();
              sum+=array[i];
          }
          long yo=n;
          long a1=2;
          long hh=((yo)*(yo+1))/a1;
          
          if(sum%hh!=0){
              res.append("NO \n");
          }
          else{
              long well=sum/hh;
              long ans[]= new long[n];
              int check=0;
              for(int i=1;i<n;i++){
                  long yo1=array[i-1]-array[i]+well;
                  if(yo1<=0){
                      check=1;
                      break;
                  }
                  else{
                      if(yo1%n!=0){
                          check=1;
                          break;
                      }
                      else{
                          long kk=yo1/n;
                          ans[i]=kk;
                      }
                  }
  
              }
              long q=array[n-1]-array[0]+well;
              if(q<=0){
                  check=1;
              }
              if(q%n!=0){
                  check=1;
              }
              else{
                  long kk=q/n;
                  ans[0]=kk;
              }
              if(check==0){
                  res.append("YES \n");
                  for(int i=0;i<n;i++){
                      res.append(ans[i]+" ");
                  }
                  res.append(" \n");
              }
              else{
                  res.append("NO \n");
              }
  
          }
          p++;
      }
      System.out.println(res);
      }
      static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
   
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
   
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
   
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
   
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
   
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
   
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  
      static long modpower(long x, long y, long p)
       {
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
  
    
     // SIMPLE POWER FUNCTION=>
  
  
        static long power(long x, long y)
          {
              long res = 1;     // Initialize result
          
              while (y > 0)
              {
              
                  // If y is odd, multiply x with result
                  if ((y & 1) != 0)
                      res = res * x;
          
                  // y must be even now
                  y = y >> 1; // y = y/2
                  x = x * x;  // Change x to x^2
              }
              return res;
          }
  
  
  
  
  
  }