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.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.Arrays;
  import java.util.StringTokenizer;
  
  public class Problem1618E {
  
  
      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;
          }
      }
  
  
      public static void main(String[] args){
          FastReader fr = new FastReader();
          int tests = fr.nextInt();
          while(tests > 0){
  
              int n = fr.nextInt();
              long[] r = Arrays.stream(fr.nextLine().split(" ")).mapToLong(Long::parseLong).toArray();
  
              long sum = Arrays.stream(r).sum();
              long k;
              long seq = (long) n * (n+1) / 2;
              if(sum % seq == 0) {
                  k = sum / seq;
              } else {
                  System.out.println("NO");
                  tests--;
                  continue;
              }
  
  
              long[] a = new long[n];
              boolean hasSol = true;
  
              for(int i=1; i<n; i++){
                  long temp = r[i-1] - r[i] + k;
                  if(temp <= 0 || temp % n != 0) {
                      hasSol = false;
                      break;
                  }
                  a[i] = temp / n;
              }
  
              if(hasSol){
                  a[0] = r[0];
                  for(int i=1; i<n; i++){
                      a[0] -= a[i]*(n+1-i);
                  }
                  if(a[0] > 0) {
                      System.out.println("YES");
                      for(int i=0; i<n; i++){
                          System.out.print(a[i]+" ");
                      }
                      System.out.println();
                  }
                  else System.out.println("NO");
              } else {
                  System.out.println("NO");
              }
  
              tests--;
          }
      }
  }
  