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.*;
import java.util.*;

public class C {
    public static Scanner sc = new Scanner(System.in);

    public static int n,k,t;
    public static long[] a;

    public static void reverseArray(long[] A){
        int n = A.length;
        long temp;
        for (int i=0; i<n/2; i++){
            temp = A[i];
            A[i] = A[n-i-1];
            A[n-i-1] = temp;
        }
    }

    public static void Input(){
        t = sc.nextInt();
    }

    public static void Solve(){
        for (int i=0; i<t; i++) {
            n = sc.nextInt();
            k = sc.nextInt();
            a = new long[n];

            for (int j = 0; j < n; j++) {
                a[j] = sc.nextLong();
            }

            Arrays.sort(a);

            int first = -1;

            for (int j = 0; j < n; j++) {
                if (a[j] != 0) {
                    first = j;
                    break;
                }

            }

            if (first == -1) {
                System.out.println("YES");
                continue;
            }

            //reverseArray(a);
            long max=1;

            while(max<=a[n-1]){
                max = max*k;
            }

            max = max/k;
            //System.out.println(max);


            for (long ex = max; ex >=1 ; ex = ex/k) {
                for (int j = n-1; j >= first; j--) {
                    if (a[j] >= ex && (a[j]/k)  < ex) {
                        a[j] = a[j] - ex;
                        //System.out.println(a[j]);
                        break;
                    }
                }
            }

            boolean flag = true;

            for (int j=0; j<n; j++){
                if (a[j]!=0){
                    System.out.println("NO");
                    flag = false;
                    break;
                }
            }

            if (flag)
                System.out.println("YES");
        }
    }

    public static void main(String args[]){
        Input();
        Solve();
    }

}
