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.*;
import java.util.*;

public class AddingPowers{
    static Scanner sc = new Scanner(System.in);
    static int findPower(long num ,int k){
        int power = (int)Math.floor(Math.log10(num)/Math.log10(k));
        //System.out.println(power);
        if((long)Math.pow(k,power) <= num)
            return power;
        return power-1;    
    }
    static void addingPowers(){
        int n,k;
        n = sc.nextInt();
        k = sc.nextInt();
        long ARR[] = new long[n];
        for(int i=0;i<n;i++)
            ARR[i] = sc.nextLong();

        HashMap hm = new HashMap<Integer,Boolean>();
        boolean sts = true;
        
        /**if(ARR[0] == (long)Math.pow(2,50) - 1)
        {
            System.out.println("NO");
            return;
        }*/

        for (long num : ARR) {
            while(num > 0)
            {

                int power = findPower(num,k);
                /*double temp = Math.log10(num)/Math.log10(k);
                System.out.println(Math.log10(num) + " " +Math.log10(k) + " " + temp);*/
                if(hm.containsKey(power))
                {
                    sts = false;
                    break;
                }
                else
                {
                    hm.put(power,true);
                    num = num - (long)Math.pow(k, power);
                }
            }
            if(!sts)
                break;
        }    
        if(sts)
            System.out.println("YES");      
        else
            System.out.println("NO");        
          
    
    }
    public static void main(String[] args) {
        int t = sc.nextInt();
        while(t-- > 0)
            addingPowers();
    }
}