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. 
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(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();
	}

}