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.
//package CodeForces;

import java.util.HashMap;
import java.util.Scanner;

public class AddingPowers {
	
	static String solve(long n,int k,HashMap<Long,Boolean> map)
	{
		if(k==0)
		{
			return "";
		}
		String res="";
		int count=0;
		while(n>0)
		{
			if(n%k==0)
			{
				res="0"+res;
			}
			else if(n%k==1)
			{
				res="1"+res;
				if(map.containsKey((long)(Math.pow(k,count))))
				{
					return "";
				}
				else
				{
					map.put((long)Math.pow(k,count),true);
				}
						
			}
			else
			{
				return "";
			}
			count++;
			n=n/k;
		}
		return res;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
		int t=s.nextInt();
		while(t>0)
		{
			int n=s.nextInt();
			int k=s.nextInt();
			boolean done=true;
			HashMap<Long,Boolean> map=new HashMap<>();
			for(int i=0;i<n;i++)
			{
				long temp=s.nextLong();
				if(temp!=0)
				{
					String str=solve(temp,k,map);
					if(str.length()==0)
					{
						done=false;
					}
				}
			}
			
			if(done)
			{
				System.out.println("YES");
			}
			else
			{
				System.out.println("NO");
			}
			
			t--;
		}

	}

}
