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.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Kraken7
 */
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();
    solver.solve(1, in, out);
    out.close();
  }

  static class TaskC {

    private int k;

    public void solve(int testNumber, Scanner in, PrintWriter out) {
      int t = in.nextInt();
      while (t-- > 0) {
        int n = in.nextInt();
        k = in.nextInt();
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
          a[i] = in.nextLong();
        int b = logk((long) 1e16);
        boolean fine = true;
        for (long i : a) {
          if (i > 1 && i < k) {
            fine = false;
            break;
          }
        }
        while (b >= 0) {
          int cnt = 0;
          long v = (long) Math.pow(k, b);
          for (int i = 0; i < n; i++) {
            if (a[i] >= v) {
              cnt++;
              a[i] -= v;
            }
          }
          if (cnt > 1) {
            fine = false;
            break;
          }
          b--;
        }
        for (long i : a) {
          if (i != 0) {
            fine = false;
            break;
          }
        }
        out.println(fine ? "YES" : "NO");
      }
    }

    private int logk(long x) {
      return (int) (Math.log10(x) / Math.log10(k));
    }

  }
}

