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.*;
  import java.io.*;
  public class a {
      static FastReader sc = new FastReader();
      static PrintWriter out = new PrintWriter(System.out);
      public static void main (String[] args) throws java.lang.Exception {        
          int t = sc.nextInt();        
          while(t-->0){  
              int  n = sc.nextInt();            
              int arr[] = new int[n];
              long sum = 0 ;
              for(int i = 0 ;i < n; i++){
                  arr[i] = sc.nextInt();
                  sum += arr[i];
              }
              long div = (long)(n*(n+1)/2);
              if(sum % div != 0){
                  out.println("NO");
                  continue;
              }
              sum /= div;
              boolean flag = true;
              long ans[] = new long[n];
              for(int i = 0 ;i < n; i++){
                  long res = arr[i]-arr[(i+1)%n]+sum;
                  if(res % n != 0 || res <= 0){
                      out.println("NO");
                      flag = false;
                      break;
                  }
                  ans[(i+1)%n] = res/n;
              }
              if(flag){
                  out.println("YES");
                  for(int i = 0 ; i < n; i++){
                      out.print(ans[i]+" ");
                  }
                  out.println();
              }
          }
          // discipline is doing what needs to be done even if you don't want to do it.
          out.flush();
      }
      
  }
  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; 
          } 
      }