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.*;
public class Solution
{
    public static void main(String[] args)
    {
        Scanner scan=new Scanner(System.in);
        int n=0,k=0,t=0;
        if(scan.hasNext())
            t=scan.nextInt();
        for(int i=0;i<t;i++) {
            if (scan.hasNext())
                n = scan.nextInt();
            if (scan.hasNext())
                k = scan.nextInt();
            long y=0;

            long[] arr=new long[n];

            for(int j=0;j<n;j++) {
                if (scan.hasNext())
                    arr[j]= scan.nextLong();
            }
            Arrays.sort(arr);
            long l=0;
            int p=0;
            List<Long> list=new ArrayList<>();
            while(l< 1e16){
                l=(long)Math.pow(k,p);
                list.add(l);
                p++;
            }
            HashSet<Long> set=new HashSet<>();
            boolean b=false;
            for(int j=0;j<n;j++){
                b=false;
                for(int u=list.size()-1;u>=0;u--){
                    if(arr[j]==0){
                        b=true;
                        break;
                    }else if(arr[j]>=list.get(u) && !set.contains(list.get(u))){
                        arr[j]-=list.get(u);
                        set.add(list.get(u));
                    }

                }
                if(arr[j]==0){
                    b=true;
                }
                if(!b){

                    break;
                }
            }
            if(!b){
                System.out.println("NO");
            }else{
                System.out.println("YES");
            }
        }

    }
}



