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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
import java.util.Comparator;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class TaskC {
        Set<Integer> used = new HashSet<>();

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            used.clear();
            int n, k;
            n = in.nextInt();
            k = in.nextInt();
            List<Long> a = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                long c = in.nextLong();
                if (c != 0) {
                    a.add(c);
                }
            }
            a.sort(Comparator.comparingInt(Long::intValue));
            for (Long i : a) {
                if (!possible(i, k)) {
                    out.println("NO");
                    return;
                }
            }
            out.println("YES");
        }

        private boolean possible(long a, int k) {
            while (a != 0) {
                int mx = findMax(a, k);
                if (mx == -1 || used.contains(mx)) {
                    return false;
                }
                used.add(mx);
                a -= (long) Math.pow(k, mx);
            }
            return true;
        }

        private Integer findMax(long a, int k) {
            int result = -1;
            for (int i = 0; (long) Math.pow(k, i) <= a; i++) {
                result = i;
            }
            return result;
        }

    }
}

