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 MyClass {
    public static void main(String args[]) {
      Scanner in=new Scanner(System.in);
      int t=in.nextInt();
      while(t-->0){
          int n=in.nextInt();
          int K=in.nextInt();
          long a[]=new long[n];
          ArrayList<Long> array=new ArrayList<>();
          for(int i=0;;i++)
          {
              long temp=(long)Math.pow(K,i); if(temp>1000000000000000000L) break; else array.add(temp);
          }
          boolean visited[]=new boolean[array.size()];
         for(int i=0;i<n;i++) a[i]=in.nextLong();
         Arrays.sort(a);
         int flag=0;
         for(int i=0;i<n;i++)
         {
            if(a[i]==0) continue; int j=0;
            for(j=0;j<array.size();j++)
            {
                if(array.get(j)>a[i]) break;
            }
            for(int k=j;k>=0;k--)
            {
                if(a[i]==0) break;
                if(array.get(k)>a[i]) continue;
                if((a[i]-array.get(k))>=0&&((a[i]-array.get(k))%K==0||(a[i]-array.get(k))%K==1)){ if(visited[k]==true) {flag=1; break; } visited[k]=true; a[i]-=array.get(k); }
            } if(flag==1) break; if(a[i]!=0) { flag=1; break; }
         }
         if(flag==1) System.out.println("NO"); else System.out.println("YES");
      }
    }
}