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 Main {
     // For fast input output
     static class FastReader {
         BufferedReader br;
         StringTokenizer st;
         public FastReader()
         { try {br = new BufferedReader(
         new FileReader("input.txt"));
         PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
         System.setOut(out);}
         catch(Exception e) { 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;
         }
     }
      // end of fast i/o code
      public static void main(String[] args) {
          FastReader reader = new FastReader();
          StringBuilder sb = new StringBuilder("");
          
          int t = reader.nextInt();
          while(t-->0){
              int n = reader.nextInt();
              
              long arr[] = new long[n];
              long ans[] = new long[n];
              boolean possible = true;
              long sum = 0;
              for(int i=0; i<n; i++){
                  arr[i] = reader.nextInt();
                  sum += arr[i];
              }
  
              if(n==1){
                  sb.append("YES\n");
                  sb.append(arr[0]+"\n");
                  continue;
              }
  
              long t1 = (1L*n*(n+1))/2;
              if(sum%t1!=0){
                  possible = false;
                  sb.append("NO\n");
                  continue;
              }
  
              t1 = sum/t1;
              // sb.append(t+" "+t1);
              for(int i=0; i<n; i++){
                  int j = i-1;
                  if(i==0){
                      j = n-1;
                  }
                  ans[i] = arr[j] - (arr[i]-t1);
                  // sb.append(t+" "+ans[i]+" "+i+" \n");
                  if(ans[i]<=0){
                      possible = false;
                      break;
                  }
                  if(ans[i]%n!=0){
                      possible = false;
                      break;
                  }
                  ans[i] /= n;
              }
  
              if(!possible){
                  sb.append("NO\n");
              }else{
                  sb.append("YES\n");
                  for(int i=0; i<n; i++){
                      sb.append(ans[i]+" ");
                  }
                  sb.append("\n");
              }
          }
          System.out.println(sb);
      }
  }