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 Main
{
	static class InputReader {
		BufferedReader reader;
		StringTokenizer tokenizer;
		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}
		String next() { // reads in the next string
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() { // reads in the next int
			return Integer.parseInt(next());
		}
		public long nextLong() { // reads in the next long
			return Long.parseLong(next());
		}
		public double nextDouble() { // reads in the next double
			return Double.parseDouble(next());
			}
		}
	static InputReader r = new InputReader(System.in);
	static PrintWriter pw = new PrintWriter(System.out);
	public static void main(String[] args)
	{
		int t = r.nextInt();
		while (t > 0)
		{
			t--;
			int n = r.nextInt();
			long[] b = new long[n];
			long[] a = new long[n];
			long co = (long)n*((long)n+1)/2;
			long sum = 0;
			for (int i = 0; i < n; i ++)
			{
				b[i] = r.nextLong();
				sum += b[i];
			}
			if (sum%co != 0)
			{
				pw.println("NO");
				continue;
			}
			sum/=co;
			boolean flag = false;
			for (int i = 0; i < n; i ++)
			{
				long x = sum - (b[i]-b[(i-1+n)%n]);
				if (x%n != 0 || x == 0)
				{
					flag = true;
				}
				else
				{
					a[i] = x/n;
				}
			}
			if (t == 1)
			{
				pw.println(sum);
			}
			if (flag)
			{
				pw.println("NO");
				continue;
			}
			pw.println("YES");
			for (int i = 0; i < n; i ++)
			{
				pw.print(a[i] + " ");
			}
			pw.println();
		}
		pw.close();
	}
	static Integer[] reverse(Integer[] a, int n)
	{
		Integer[] b = new Integer[n];
		for (int i = 0; i < n; i ++)
		{
			b[i] = a[n-i-1];
		}
		return b;
	}
	static long gcd(long a, long b) {
   if (b==0) return a;
   return gcd(b,a%b);
}
}