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 sing {
     public static void main(String[] hi) {
        FastIO io = new FastIO();
        int t = io.nextInt();
        while (t --> 0) {
           int n = io.nextInt();
           int[] array = new int[n];
           long sum = 0;
           for (int i=0; i<n; i++) {
              array[i] = io.nextInt();
              sum += array[i];
           }
           long permutation = (n * (n+1))/2;
           if (sum % permutation != 0) {
              io.println("NO");
           }
           else {
              long temp = Long.valueOf(sum/permutation);
              boolean solved = true;
              ArrayList<Long> answer = new ArrayList<Long>();
              long x = array[n-1] - array[0] + temp;
              if (x%n != 0) {
                 solved = false;
              }
              x /= n;
              if (x <= 0) {
                 solved = false;
              }
              answer.add(x);
              for (int i=1; i<n; i++) {
                 int curr = array[i];
                 int prev = array[(i+n-2)%(n-1)];
                 long y = prev - curr + temp;
                 if (y % n != 0) {
                    solved = false;
                 }
                 y /= n;
                 if (y <= 0) {
                    solved = false;
                 }
                 answer.add(y);
              }
              if (solved) {
                 io.println("YES"); 
                 for (long i : answer) {
                    io.print(i + " ");
                 }
                 io.println();
              }
              else {
                 io.println("NO");
              }
           }
        }
        io.close();
     }
     static class FastIO extends PrintWriter {
        BufferedReader br;
        StringTokenizer st;
        public FastIO() {
           super(System.out);
           br = new BufferedReader(new InputStreamReader(System.in));
        }
        public String next() {
           try {
              while (st == null || !st.hasMoreTokens()) {
                 st = new StringTokenizer(br.readLine());
              }
              return st.nextToken();
           } 
           catch (Exception e) {
              return null;
           }
        }
        public int nextInt() {
           return Integer.parseInt(next());
        }
        public double nextDouble() {
           return Double.parseDouble(next());
        }
        public long nextLong() {
           return Long.parseLong(next());
        }
        public void close() {
           try {
              br.close();
           } 
           catch (IOException e) {
              e.printStackTrace();
           }
           super.close();
        }
     }
  }