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.*;
  import java.util.*;
  
  public class new1{
  
  	public static void main(String[] args)  throws IOException{
      	//BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
  		FastReader s = new FastReader();
  		int t = s.nextInt();
      	for(int z = 0; z < t; z++) {
      		int n = s.nextInt();
      		int[] arr = new int[n];
      		long sum = 0;
      		for(int i = 0; i < n; i++) {
      			arr[i] = s.nextInt();
      			sum = sum + arr[i];
      		}
      		if(sum % (((long) n * (n + 1)) / 2) != 0) {
      			System.out.println("NO");
      			continue;
      		}
      		sum = sum / (((long) n * (n + 1)) / 2);
  
      		long[] ans= new long[n];
      		int pos = 1;
      		//System.out.println(sum);
      		for(int i = 0; i < n; i++) {
      			int j = (i - 1);
      			if(j < 0) j = n - 1;
      			//long ab = (arr[i] - arr[j]);
      			ans[i] = (sum - arr[i] + arr[j]) / n;
  //    			System.out.println(i + " " + (ab));
      			if((sum - arr[i] + arr[j]) % n != 0) pos = 0;
      			if(ans[i] <= 0 || ans[i] > 1000000000) pos = 0;
      		}
      		if(pos == 0) System.out.println("NO");
      		else {
      			System.out.println("YES");
      			for(int i = 0; i < n; i++) System.out.print(ans[i] + " ");
      			System.out.println();
      		}
      		
      	}
  //    	output.flush();
  
      }
  }
  
  
  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();
      }
   
      public 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;
  }}