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 E {
	
	public static void main(String[] args)throws IOException {
		
		FastScanner scan = new FastScanner();
		PrintWriter output = new PrintWriter(System.out);
		int t = scan.nextInt();
		for(int tt = 0;tt<t;tt++) {
			int n = scan.nextInt();
			int arr[] = scan.readArray(n);
			long sum = 0;
			for(int i:arr) sum+=i;
			long div = (n*(n+1))/2;
			if(sum%div!=0) {
				output.println("NO");
				continue;
			}
			long ans[] = new long[n];
			sum/=div;
			
			boolean possible = true;
			for(int i = 0;i<n;i++) {
				long num = arr[i]-arr[(i+1)%n]+sum;
				if(num<=0 || num%n!=0) {
					possible = false;
					break;
				}
				ans[(i+1)%n] = num/n;
			}
			if(!possible) {
				output.println("NO");
				continue;
			}
			output.println("YES");
			for(long i : ans) output.print(i+" ");
			output.println();
			
		}
		output.flush();
	}
	
	public static int[] sort(int arr[]) {
		List<Integer> list = new ArrayList<>();
		for(int i:arr) list.add(i);
		Collections.sort(list);
		for(int i = 0;i<list.size();i++) arr[i] = list.get(i);
		return arr;
	}
	
	public static int gcd(int a, int b) {
		if(a == 0) return b;
		return gcd(b%a, a);
	}
	
	public static void printArray(int arr[]) {
		for(int i:arr) System.out.print(i+" ");
		System.out.println();
	}
	
	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;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}

}
