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.io.*;
import java.util.*;

public class new1{
	
    public static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
        return gcd(b%a, a);
    }
	
	

	public static void main(String[] args)  throws IOException{
    	BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
		FastReader s = new FastReader();
		int t = s.nextInt();
    	for(int z = 0; z < t; z++) {
    		int n = s.nextInt();
    		int[] arr = new int[n];
    		long sum = 0;
    		for(int i = 0; i < n; i++) {
    			arr[i] = s.nextInt();
    			sum = sum + arr[i];
    		}
    		sum = sum / ((n * (n + 1)) / 2);
    		//System.out.println(sum);
    		long[] ans= new long[n];
    		int pos = 1;
    		for(int i = 0; i < n; i++) {
    			int j = (i - 1);
    			if(j < 0) j = n - 1;
    			long aa = sum - arr[i] + arr[j];
    			
    			ans[i] = (sum - arr[i] + arr[j]) / n;
    		//	System.out.println(ans[i] + " " + (sum - arr[i] + arr[j]) + " " + i);
    			if((sum - arr[i] + arr[j]) % n != 0) pos = 0;
    			if(ans[i] == 0) pos = 0;
    			
    		}
    		if(pos == 0) System.out.println("NO");
    		else {
    			System.out.println("YES");
    			for(int i = 0; i < n; i++) System.out.print(ans[i] + " ");
    			System.out.println();
    		}
    		
    	}
//    	output.flush();

    }
}


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();
    }
 
    public 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;
}}