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.util.*;
import java.io.*;
public class Solution{
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out=new PrintWriter(System.out);
		String[] temp=br.readLine().trim().split(" ");
		int numTestCases=Integer.parseInt(temp[0]);
		while(numTestCases-->0){
		    temp=br.readLine().trim().split(" ");
		    int n=Integer.parseInt(temp[0]);
		    int k=Integer.parseInt(temp[1]);
		    temp=br.readLine().trim().split(" ");
		    long[] arr=new long[n];
		    for(int i=0;i<n;i++){
		        arr[i]=Long.parseLong(temp[i]);
		    }
		    out.println(isPossible(arr,k));
		}
		out.flush();
		out.close();
	}
	public static String isPossible(long[] arr,int k){
	    int power=(int)(Math.log(10000000000000000L)/Math.log(k));
	    while(power>=0)
	    {
	        for(int i=0;i<arr.length;i++){
    	        if((long)Math.pow(k,power)<=arr[i]){
    	            arr[i]-=(long)Math.pow(k,power);
    	            break;
    	        }
    	    }
    	    power--;
	    }   
	    for(int i=0;i<arr.length;i++){
	        if(arr[i]!=0){
	            return "NO";
	        }
	    }
	    return "YES";
	}
}