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 class Main {
    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int T;
        int[] vis=new int[60];
        int n;
        int flag;
        long k;
        long number;
        T=in.nextInt();
        for(int i=1;i<=T;i++)
        {
            for(int a=0;a<60;a++){
                vis[a]=0;
            }
            flag=1;
            n = in.nextInt();
            k = in.nextLong();
            for (int j = 1; j <= n; j++) {
                number = in.nextLong();
                for(int b=1;;b++) {
                    if (number == 0) break;
                    if (number % k != 0) {
                        if ((number - 1) % k != 0) {
                            flag = 0;
                            break;
                        } else {
                            number--;
                            if(vis[b-1]==0){
                                vis[b-1]=1;
                            } else {
                                flag=0;
                                break;
                            }
                            number /= k;
                        }
                    } else {
                        number /= k;
                    }
                }
            }
            if(flag==1){
                System.out.println("YES");
            }else {
                System.out.println("NO");
            }
        }
    }
}
