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.Scanner;
  
  public class eqN {
      static void printArr(long[] a){
          for(int i=0; i<a.length; i++){
              System.out.print(a[i]+" ");
          }
          System.out.println();
      }
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int t= sc.nextInt();
          while(t-->0){
              int n= sc.nextInt();
              long[] b= new long[n];
              long sum = 0;
              for(int i=0; i<n; i++) {
                  b[i]= sc.nextLong();
                  sum+=b[i];
              }
              //printArr(b);
              if(n==1){
                  System.out.println("YES");
                  System.out.println(b[0]);
                  continue;
              }
              if(n==2){
                  if(sum%3==0&&sum>3){
                      sum/=3;
                      if(b[0]-sum>0&b[1]-sum>0){
                          System.out.println("YES");
                          System.out.println((b[1]-sum)+" "+(b[0]-sum));
                      }
                      else{
                          System.out.println("NO");
                      }
              }
                  else
                  {
                      System.out.println("NO");
                  }
                  continue;
              }
              long[] ar = new long[n-1];
              for(int i=0; i<n-1; i++) {
                  ar[i]=b[i+1]-b[i];
              }
             //printArr(ar);
              long tmp=0;
              int tmp2= n*(n+1)/2;
              long[] ans = new long[n];
              //System.out.println();
              //System.out.println(sum);
              long p=0;
              boolean d= false;
              if(sum%tmp2==0&&sum>tmp2){
                  sum/=tmp2;
                  //System.out.println("PPPP");
                  for(int i=0; i<n-1; i++) {
                      p=((sum-ar[i])/n);
                      //System.out.println(p);
                      if(p>0){
                      ans[i+1]=p;
                      d=true;
                      }
                      else{
                          d=false;
                          break;
                      }
                      tmp+=ans[i+1];
                  }
                  ans[0]=(sum-(b[0]-b[n-1]))/n;
                  tmp+=ans[0];
                  if(tmp!=sum||ans[0]<=0)
                      d=false;
              }
              if(d){
                  System.out.println("YES");
                  printArr(ans);
              }
              else{
                  System.out.println("NO");
              }
  
          }
      }
  }
  