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 path {

	static boolean check(long d, long[] arr) {
		if (d == 0)
			return true;
		for (int i = 1; i < arr.length; i++) {
			if (arr[i - 1] % d == 0 && arr[i] % d == 0)
				return false;
			if (arr[i - 1] % d != 0 && arr[i] % d != 0)
				return false;
		}
		return true;
	}

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

	/*
	 * static long binarySearch(long start, long end, long [] arr){ if(arr.length ==
	 * 2) return Math.max(arr[0], arr[1]);
	 * 
	 * while(start < end){ long mid = (start + end) / 2; if(check(mid, arr)){ ans =
	 * mid; start = mid + 1; } else end = mid; }
	 * 
	 * for(int i = 0; i < arr.length - 2; i += 2) { long p = gcd(arr[i], arr[i +
	 * 2]); if(p != 1) { if(check(p, arr)) return p; } } for(int i = 1; i <
	 * arr.length - 2; i += 2) { long p = gcd(arr[i], arr[i + 2]); if(p != 1) {
	 * if(check(p, arr)) return p; } } return 0; }
	 */
	
	public static void main(String[] args) throws IOException {
		BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
		int q = Integer.parseInt(sc.readLine());
		for (int t = 0; t < q; t++) {
			int n = Integer.parseInt(sc.readLine());
			String [] s = sc.readLine().split(" ");
			int [] arr = new int [n];
			int sum = 0;
			for(int i = 0; i < n; i++) {
				arr[i] = Integer.parseInt(s[i]);
				sum += arr[i];
			}
			int pre = (n * (n + 1)) / 2;
			if(sum % pre == 0 && sum / pre >= n) {
				System.out.println("YES");
				int temp = 0;
				for(int i = 0; i < n - 1; i++) {
					System.out.print(1 + " ");
					temp++;
				}
				System.out.println((sum / pre) - temp);
			}
			else {
				System.out.println("NO");
			}
		}
	}
}