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 Main {
	public static PrintWriter out;
	public static InputReader in;

	public static void main(String[] args)throws IOException {
		in = new InputReader(System.in);
		out = new PrintWriter(System.out);

		int t = in.nextInt();
		int n,k;
		long arr[];
		for(int x=0;x<t;x++)
		{
			n=in.nextInt();
			arr=new long[n];
			k=in.nextInt();
			long mx=-1;
			// Arrays.fill(ct,0);
			for(int i=0;i<n;i++){
				arr[i]=in.nextLong();
				mx=Math.max(mx,arr[i]);
			}
			if(mx==0)
			{
				out.println("YES");
				continue;
			}
			int mxpw = (int)(Math.log(mx)/Math.log(k))+1;
			// out.println("mxpw "+mxpw);
			int ct[] = new int[mxpw+1];
			Arrays.fill(ct,0);
			// long val = (long) Math.pow(k,mxpw);
			for(int i=0;i<n;i++)
			{
				long val = (long) Math.pow(k,mxpw);
				for(int j=mxpw;j>=0;j--)
				{
					if(val<=arr[i])
					{
						ct[j]++;
						arr[i]-=val;
						
					}
					val=val/k;
				}
			}
			boolean ans = true;
			for(int i=0;i<=mxpw;i++)
			{
				if(ct[i]>1)
				{
					ans=false;
					break;
				}
			}
			for(int i=0;i<n;i++)
			{
				if(arr[i]>0)
					ans=false;
			}
			if(ans)
				out.println("YES");
			else
				out.println("NO");
		}
		out.close();
	}
	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}