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 prac {
  static Scanner sc = new Scanner(System.in);
  static long mod=1000000007;
  public static void main(String args[]) {
    long t=sc.nextLong();
   while(t-->0)
   {
   int n=sc.nextInt(),k=sc.nextInt();
   long a[]=new long[n];
   long max=0;
   for(int i=0;i<n;i++)
   {
   a[i]=sc.nextLong();
   max=Math.max(a[i],max);
   }
   long b[]=new long[n];
   long z=0;
   long ind=0;
   while(max>z)
   {
     ind++;
     z=(long)Math.pow(k, ind);
   }
   for(long i=ind;i>=0;i--)
   {
     long t1=(long)Math.pow(k, i);
     for(int j=0;j<n;j++)
     {
       if(a[j]>=t1)
       {
         a[j]-=t1;
         break;
        //  b[j]+=t;  
       }
     }
   }
   boolean isPoss=true;
   for(int i=0;i<n;i++)
   {
     if(a[i]!=0)
     {
       System.out.println("NO");
       isPoss=false;
       break;
      }
    }
    if(isPoss)
    System.out.println("YES");
   }
     }
}
