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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;

public class problem {
	static long  [] bin=new long[100];
	
 	static void de(long a,int k)
 	{
 		int i=0;
 		
 		while(a>0)
 		{
 			bin[i]+= (a%k);
 			a=a/k;
 			i++;
 		}
 	}
	
		public static void main(String[] args) {
	
		Scanner scan=new Scanner(System.in);
		
		int t=scan.nextInt();
		
		
		while(t>0)
		{
			
		int n=scan.nextInt();
		int k=scan.nextInt();
		long [] a=new long[n];
		int i;
				
		Arrays.fill(bin, 0);
		for(i=0;i<n;i++) {
		a[i]=scan.nextLong();
		de(a[i],k);
	
		}
	boolean p=true;
		for(i=0;i<100;i++)
		{
			if(bin[i]>1) {
				p=false;
				break;
			}
				
		}
		
		if(p)
			System.out.println("YES");
		else
		System.out.println("NO");
		
			t--;
		}

	}

}
