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: // package codeforces;
  import java.io.*;
  import java.util.*;
  	
  public class practice {
  	
  //	static PrintWriter out  = new PrintWriter(System.out);
  //	static StringBuilder sb = new StringBuilder();
  	static FastScanner fs = new FastScanner();
  		
  	public static void main(String[] args) {
  		int t = 1;
  		t = fs.nextInt();
  		for(int i=1;i<=t;i++) {
  			solve(t);
  //			out.println(sb);
  //			out.close();
  		}
  	}
  		
  	public static void solve(int t) {
  		int n = fs.nextInt();
  		long b[] = new long[n];
  		long sum=0;
  		for(int i=0;i<n;i++) {
  			b[i]=fs.nextLong();sum+=b[i];
  		}
  		long z = ((long)n)*(((long)n)+1)/2;
  		if(sum%z != 0) {
  			System.out.println("NO");
  			return;
  		}
  		long k = sum/z;
  		long a[] = new long[n];
  		for(int i=0;i<n;i++) {
  			long x = k + b[(n + i - 1) % n]  - b[i];
  			if(x%n!=0 || x<=0) {
  				System.out.println("NO");return;
  			}
  			a[i]=x/n;
  		}
  		System.out.println("YES");
  		for(int i=0;i<n;i++) {
  			System.out.print(a[i]+" ");
  		}
  		System.out.println();
  		return;
  	}
  		
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  			
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		ArrayList<Integer> readList(int n) {
  			ArrayList<Integer> a = new ArrayList<Integer>();
  			int x;
  			for(int i=0;i<n;i++) {
  				x=fs.nextInt();
  				a.add(x);
  			}
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  }
  