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.io.BufferedInputStream;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(new BufferedInputStream(System.in));
        int t=sc.nextInt();
        while(t-->0){
            int v=0;
            int n=sc.nextInt();
            long m[]=new long[n+1];
            int a[]=new int[200];
            int x=sc.nextInt();
            for(int i=1;i<=n;i++){
                int count=0;
                m[i]=sc.nextLong();
                if(v==0) {
                    while ((m[i] % x == 0 || m[i] % x == 1) && m[i] != 0) {
                        count++;
                        if (m[i] % x == 1) {
                            m[i]--;
                            a[count - 1]++;
                            if (a[count - 1] == 2) {
                                v++;
                                break;
                            }
                        }
                        m[i] /= x;
                    }
                    if (m[i] != 0)
                        v++;
               }
           }
            if(v!=0)
                System.out.println("NO");
            else
                System.out.println("YES");
       }
    }
}