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.util.*;
  import java.io.*;
  
  public class B {
  	static 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();
  		}
  
  		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;
  		}
  	}
  
  	public static void sort(long[] a) {
  		ArrayList<Long> arr = new ArrayList<>();
  		for (int i = 0; i < a.length; i++)
  			arr.add(a[i]);
  		Collections.sort(arr);
  		for (int i = 0; i < arr.size(); i++)
  			a[i] = arr.get(i);
  	}
  
  	public static void revsort(long[] a) {
  		ArrayList<Long> arr = new ArrayList<>();
  		for (int i = 0; i < a.length; i++)
  			arr.add(a[i]);
  		Collections.sort(arr, Collections.reverseOrder());
  		for (int i = 0; i < arr.size(); i++)
  			a[i] = arr.get(i);
  	}
  
  	//Cover the small test cases like for n=1 .
  
  	public static void main(String[] args) {
  		FastReader obj = new FastReader();
  		PrintWriter out = new PrintWriter(System.out);
  		int len = obj.nextInt();
  		while (len-- != 0) {
  			int n = obj.nextInt();
  			long[] b=new long[n];
  			long[] a=new long[n];
  			long sum=n*(n+1)/2,s=0;
  			for(int i=0;i<n;i++) {
  				b[i]=obj.nextLong();
  				s+=b[i];
  			}
  			if(s%sum==0) {
               long S=s/sum;
               int k=0;
               for(int i=1;i<n;i++)
               {
              	a[i]=(b[i-1]+S-b[i]);
              	if(a[i]%n!=0 || a[i]<=0)
              	{
              		k=1;
              		break;
              	}
              	a[i]=a[i]/n;
               }
               a[0]=(b[n-1]+S-b[0])/n;
               if(a[0]<=0)k=1;
               if(k==0) {
              	 out.println("YES");
              	 for(int i=0;i<n;i++)out.print(a[i]+" ");
               }
               else out.print("NO");
               out.println();
  			}
  			else out.println("NO");
  		}
  		out.flush();
  	}
  
  }