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: // Don't place your source in a package
  
  
  import javax.swing.*;
  import java.lang.reflect.Array;
  import java.text.DecimalFormat;
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  import java.math.*;
  import java.util.stream.Stream;
  
  
  
  // Please name your class Main
  public class Main {
      static FastScanner fs=new FastScanner();
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          public String next() {
              while (!st.hasMoreElements())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
          int Int() {
              return Integer.parseInt(next());
          }
  
          long Long() {
              return Long.parseLong(next());
          }
  
          String Str(){
              return next();
          }
      }
  
  
      public static void main (String[] args) throws java.lang.Exception {
          PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
          //reading /writing file
          //Scanner in=new Scanner(System.in);
          //Scanner in=new Scanner(new File("input.txt"));
          //PrintWriter pr=new PrintWriter("output.txt");
  
  
          int T=Int();
          for(int t=0;t<T;t++){
              Solution sol1=new Solution(out,fs);
              sol1.solution(t);
          }
          out.flush();
  
      }
  
  
      public static int[] Arr(int n){
          int A[]=new int[n];
          for(int i=0;i<n;i++)A[i]=Int();
          return A;
      }
      public static int Int(){
          return fs.Int();
      }
      public static long Long(){
          return fs.Long();
      }
      public static String Str(){
          return fs.Str();
      }
  
  }
  
  
  
  
  
  
  
  
  class Solution {
      PrintWriter out;
      int INF = Integer.MAX_VALUE;
      int NINF = Integer.MIN_VALUE;
      int MOD = 998244353;
      int mod = 1000000007;
      Main.FastScanner fs;
  
      //int mu[];
      //int musum[];
  
      public Solution(PrintWriter out,Main.FastScanner fs) {
          this.out = out;
          this.fs=fs;
          //muinit(100000);
      }
  
  
      /*public void muinit(int n){
          int N = n + 10;
          mu = new int[N];
          musum = new int[N];
          mu[1] = 1;
  
  
          boolean seen[] = new boolean[N];
          int primes[] = new int[N];
          int cnt = 0;
  
          for (int i = 2; i < mu.length; i ++ ) {
              if (!seen[i]) {
                  primes[cnt++] = i;
                  mu[i] = -1;
              }
              for (int j = 0; primes[j] * i < N; j ++ ) {
                  seen[primes[j] * i] = true;
                  if (i % primes[j] == 0) break;
                  mu[primes[j] * i] = -mu[i];
              }
          }
  
          for (int i = 1; i < N; i ++ ){
              musum[i] = musum[i - 1] + mu[i];
          }
      }*/
  
  
  
  
      public void solution(int testcase) {
          int n = fs.Int();
          long B[]=new long[n];
          long sum = 0;
          for(int i = 0; i < n; i++){
              B[i]=fs.Int();
              sum += B[i];
          }
          long res[] = new long[n];
          if((sum % ((1l + n) * n / 2)) != 0){
              out.println("NO");
              return;
          }
  
          long avg = (sum / ((1l + n) * n / 2));
  
          for(int i = 0; i < n; i++){
              int j = (i + 1) % n;
              long dif = B[j] - B[i];
              dif -= avg;
              if(Math.abs(dif) % n != 0){
                  out.println("NO");
                  return;
              }
              res[j] = -(dif / n);
              if(res[j] < 1) {
                  out.println("NO");
                  return; 
              }
          }
  
  
          out.println("YES");
          for(long i:res){
              out.print(i+" ");
          }out.println();
      }
  
  }
  
  