Suppose you are performing the following algorithm. There is an array v_1, v_2, ..., v_n filled with zeroes at start. The following operation is applied to the array several times — at i-th step (0-indexed) you can: 

  * either choose position pos (1 ≤ pos ≤ n) and increase v_{pos} by k^i; 
  * or not choose any position and skip this step. 



You can choose how the algorithm would behave on each step and when to stop it. The question is: can you make array v equal to the given array a (v_j = a_j for each j) after some step?

Input

The first line contains one integer T (1 ≤ T ≤ 1000) — the number of test cases. Next 2T lines contain test cases — two lines per test case.

The first line of each test case contains two integers n and k (1 ≤ n ≤ 30, 2 ≤ k ≤ 100) — the size of arrays v and a and value k used in the algorithm.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^{16}) — the array you'd like to achieve.

Output

For each test case print YES (case insensitive) if you can achieve the array a after some step or NO (case insensitive) otherwise.

Example

Input


5
4 100
0 0 0 0
1 2
1
3 4
1 4 1
3 2
0 1 3
3 9
0 59049 810


Output


YES
YES
NO
NO
YES

Note

In the first test case, you can stop the algorithm before the 0-th step, or don't choose any position several times and stop the algorithm.

In the second test case, you can add k^0 to v_1 and stop the algorithm.

In the third test case, you can't make two 1 in the array v.

In the fifth test case, you can skip 9^0 and 9^1, then add 9^2 and 9^3 to v_3, skip 9^4 and finally, add 9^5 to v_2.

import java.io.*;
import java.util.*;
	 
public class C 
{
	 
	public static void main(String[] args)throws IOException 
	{	
		BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));
		StringBuffer sb=new StringBuffer();
		
		int test=Integer.parseInt(ob.readLine());
		while(test-->0)
		{
			
			String s[]=ob.readLine().split(" ");
			int n=Integer.parseInt(s[0]);
			int k=Integer.parseInt(s[1]);
			
			long a[]=new long[n];
			String str[]=ob.readLine().split(" ");
			for(int i=0;i<n;i++)
				a[i]=Long.parseLong(str[i]);
			
			long pow=1;
			long max=10000000000000000L;
			boolean flag=true;
			
			while(pow<max)
				pow*=k;     
			
			while(pow!=0) 
	        {
				int count=0;
	            for(int i=0;i<n;++i) 
	            {
	            	if(a[i]>=pow)
	            	{  
	            		a[i]-=pow;
	                    ++count;
	                }
	            }
	            if(count>1)
	         		flag=false;
	            
	            if(!flag) 
	            	break;
	            pow /= k;    
	        }
	         
			if(flag) 
			{
				for(int i=0;i<n;++i)
				{
					if(a[i]!=0) 
					{   
						flag=false;
						break;
	                }
					
	             }
	        }
			if(flag)     
				sb.append("YES\n");
	        else
	            sb.append("NO\n");
		}
		System.out.println(sb);
	}
}
