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: /*
    Challenge 1: Newbie to CM in 1year (Dec 2021 - Nov 2022)   🔥 5* Codechef
    Challenge 2: CM to IM in 1 year (Dec 2022 - Nov 2023)    🔥🔥 6* Codechef
    Challenge 3: IM to GM in 1 year (Dec 2023 - Nov 2024)  🔥🔥🔥 7* Codechef
    Goal: Become better in CP!
    Key: Consistency!
  */
  
  import java.util.*;
  import java.io.*;
  import java.math.*;
  
  public class Coder {
  
    static int n, k;
    static long b[];
    static StringBuffer str=new StringBuffer();
  
    static void solve(){
      long sum=0;
      long div= n*(n+1)/2;
  
      for(int i=0;i<n;i++) sum+=b[i];
      if(sum%div !=0){
        str.append("NO\n");
        return;
      }
      sum/=div;
      long a[]=new long[n];
      for(int i=0;i<n;i++){
        // str.append((i+n-1)%n).append(" ").append(i).append("\n");
        long val=b[(i+n-1)%n]-b[i]+sum;
        if(val<=0 || val%n != 0){
          str.append("NO\n");
          return;
        }
        a[i]=val/n;
      }
      str.append("YES\n");
      for(int i=0;i<n;i++) str.append(a[i]).append(" ");
      str.append("\n");
    }
  
    public static void main(String[] args) throws java.lang.Exception {
      BufferedReader bf;
      PrintWriter pw;
      boolean lenv=false;
      if(lenv){
        bf = new BufferedReader(
                            new FileReader("input.txt"));
        pw=new PrintWriter(new
              BufferedWriter(new FileWriter("output.txt")));
      }else{
        bf = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new OutputStreamWriter(System.out));
      }
      
      int t = Integer.parseInt(bf.readLine().trim());
      while (t-- > 0) {
        String []st=bf.readLine().trim().split("\\s+");
        n=Integer.parseInt(st[0]);
        b=new long[n];
        st=bf.readLine().trim().split("\\s+");
        for(int i=0;i<n;i++) b[i]=Long.parseLong(st[i]);
          solve();
      }
      pw.print(str);
      pw.flush();
      // System.outin.print(str);
    }
  }