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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.*;


 
public class Codeforces {
	
	static int mod= 1000000007;
	
	public static void main(String[] args) throws Exception {
		PrintWriter out=new PrintWriter(System.out);
		FastScanner fs=new FastScanner();
		int t=fs.nextInt();
		outer:while(t-->0) {
			int n=fs.nextInt();
			long b[]=fs.lreadArray(n);
			long sum=0;
			for(int i=0;i<n;i++) sum+=b[i];
			long a[]=new long[n];
			long nn= (n*(long)(n+1))/2;
			if(sum%nn!=0) {
				out.println("NO");
				continue outer;
			}
			long sa= sum/nn;
			for(int i=0;i<n;i++) {
				int cur=i, next= (i+1)%n;
				long num= b[cur]-b[next]+sa;
				if(num%n!=0) {
					out.println("NO");
					continue outer;
				}
				a[next]= num/n;
				if(a[next]<=0) {
					out.println("NO");
					continue outer;
				}
			}
			out.println("YES");
			for(int i=0;i<n;i++) {
				out.print(a[i]+" ");
			}
			out.println();
		}
		out.close();
		
	}
	
	static long pow(long a,long b) {
		if(b<=0) return 1;
		long res=1;
		while(b!=0) {
			if((b&1)!=0) {
				res*=a;
				res%=mod;
			}
			a*=a;
			a%=mod;
			b=b>>1;
		}
		return res;
	}
	static long gcd(long  a,long  b) {
		if(b==0) return a;
		return gcd(b,a%b);
	}
	static long nck(int n,int k) {
		if(k>n) return 0;
		long res=1;
		res*=fact(n);
		res%=mod;
		res*=modInv(fact(k));
		res%=mod;
		res*=modInv(fact(n-k)); 
		res%=mod;
		return res;
	}
	static long fact(long n) {
//		return fact[(int)n];
		long res=1;
		for(int i=2;i<=n;i++) {
			res*=i;
			res%=mod;
		}
		return res;
	}
	
	static long modInv(long n) {
		return pow(n,mod-2);
	}
	
	static void sort(int[] a) {
		//suffle
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n);
			int temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		
		//then sort
		Arrays.sort(a);
	}
	
	// Use this to input code since it is faster than a Scanner
	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());
		}
		long[] lreadArray(int n) {
			long a[]=new long[n];
			for(int i=0;i<n;i++) a[i]=nextLong();
			return a;
		}
		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());
		}
	}
 
}