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.*;

import static java.lang.Math.*;

public class C {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);

        int t = in.nextInt();
        for (int tt = 0; tt < t; tt++) {
            int n = in.nextInt();
            long k = in.nextLong();
            long[] arr = new long[n];
            Arrays.setAll(arr, i -> in.nextLong());
            pw.println(solve(arr, n, k) ? "YES" : "NO");
        }
        pw.close();
    }

    static boolean solve(long[] arr, int n, long k) {
      //  debug(arr);
        HashSet<Long> powers = new HashSet<>();
        for (long a : arr) {
            if (a != 0) {
                long power = 1;
                while (power < a) power *= k;
                while (power > 0) {
                    if (a >= power) {
                        a -= power;
                        if (!powers.add(power)) {
                            return false;
                        }
                    }
                    power /= k;
                }
                //   debug(a);
                if (a != 0) return false;
            }
        }
        return true;
    }


    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }

}
