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;

public class AddingPowers {
	public static void main(String[] args) throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int N = Integer.parseInt(br.readLine());
		long[] array = new long[30]; String[] temp; int num; int k; 
		boolean gotone; boolean possible;
		for (int i = 0; i < N; i++)
		{
		    temp = br.readLine().split(" ");
		    num = Integer.parseInt(temp[0]); k = Integer.parseInt(temp[1]);
		    temp = br.readLine().split(" ");
		    for (int a = 0; a < num; a++)
		    	array[a] = Long.parseLong(temp[a]);
		    possible = true;
		    outer: for (int a = 0; a < 55; a++) //upper bound 55
		    {
			    gotone = false;
		    	for (int b = 0; b < num; b++)
		    	{
		    		
		    		if (array[b]%k != 0)
		    		{
		    			if (array[b]%k != 1)
		    			{
		    				possible = false;
		    				break outer;
		    			}	
		    			if (gotone == false)
		    				gotone = true;
		    			else
		    			{
		    				possible = false;
		    				break outer;
		    			}
		    		}
		    		array[b] = array[b]/k;
		    	}
		    }
		    if (possible)
		    	System.out.println("YES");
		    else
		    	System.out.println("NO");
		}
	}
}
