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 Main
  {
  	static class InputReader {
  		BufferedReader reader;
  		StringTokenizer tokenizer;
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  		String next() { // reads in the next string
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				try {
  					tokenizer = new StringTokenizer(reader.readLine());
  				} catch (IOException e) {
  					throw new RuntimeException(e);
  				}
  			}
  			return tokenizer.nextToken();
  		}
  		public int nextInt() { // reads in the next int
  			return Integer.parseInt(next());
  		}
  		public long nextLong() { // reads in the next long
  			return Long.parseLong(next());
  		}
  		public double nextDouble() { // reads in the next double
  			return Double.parseDouble(next());
  			}
  		}
  	static InputReader r = new InputReader(System.in);
  	static PrintWriter pw = new PrintWriter(System.out);
  	public static void main(String[] args)
  	{
  		int t = r.nextInt();
  		while (t > 0)
  		{
  			t--;
  			int n = r.nextInt();
  			long[] b = new long[n];
  			long[] a = new long[n];
  			long co = (long)n*((long)n+1)/2;
  			long sum = 0;
  			for (int i = 0; i < n; i ++)
  			{
  				b[i] = r.nextLong();
  				sum += b[i];
  			}
  			if (sum%co != 0)
  			{
  				pw.println("NO");
  				continue;
  			}
  			sum/=co;
  			boolean flag = false;
  			for (int i = 0; i < n; i ++)
  			{
  				long x = sum - (b[i]-b[(i-1+n)%n]);
  				if (x%n != 0 || x <= 0)
  				{
  					flag = true;
  				}
  				else
  				{
  					a[i] = x/n;
  				}
  			}
  			if (flag)
  			{
  				pw.println("NO");
  				continue;
  			}
  			pw.println("YES");
  			for (int i = 0; i < n; i ++)
  			{
  				pw.print(a[i] + " ");
  			}
  			pw.println();
  		}
  		pw.close();
  	}
  	static Integer[] reverse(Integer[] a, int n)
  	{
  		Integer[] b = new Integer[n];
  		for (int i = 0; i < n; i ++)
  		{
  			b[i] = a[n-i-1];
  		}
  		return b;
  	}
  	static long gcd(long a, long b) {
     if (b==0) return a;
     return gcd(b,a%b);
  }
  }