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.Scanner;

public final class AddingPowersCF {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while (t-->0){
            int n=sc.nextInt();
            int k=sc.nextInt();
            long arr[]=new long[n];
            for(int i=0;i<n;i++){
                arr[i]=sc.nextLong();
            }
            int s=helper(k);
            boolean ans=true;
            for(int i=s;i>=0;i--){

                boolean isTrue=false;
                int index=-1;
                for(int j=0;j<n;j++){
                    if(arr[j]>=pows(k,i)){
                        if(!isTrue){
                            isTrue=true;
                            index=j;
                        }else{
                            ans=false;
                            break;
                        }
                    }else
                        continue;
                }
                if(!ans)
                    break;

                if(isTrue){
                   // System.out.println(index+" "+i+" "+pows(k,i));
                    arr[index]-=pows(k,i);
                }

            }

            /*for(long a:arr)
                System.out.print(a+" ");
            System.out.println();*/
            if(ans && isConverted(arr,n) ) {
                System.out.println("YES");
            }
            else
                System.out.println("NO");
        }
    }

    private static long pows(int k, int i) {

        if(i==0)
            return 1;

        long ans=k;
        for(int a=1;a<i;a++){
            ans=ans*k;
        }

        return ans;
    }

    private static boolean isConverted(long[] arr, int n) {

        for(int i=0;i<n;i++){
            if(arr[i]>0)
                return false;
        }
        return true;
    }

    private static int helper(int k) {

        int low=0;
        int high=55;
        int ans=0;
        while (low<=high){
            int mid=low+(high-low)/2;
            if(Math.pow(k,mid)>=Math.pow(10,16)){
                ans=mid;
                high=mid-1;
            }else{
                low=mid+1;
            }
        }
        return  ans;
    }
}
