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.ArrayList;
import java.util.Comparator;
import java.util.HashSet;
import java.util.Scanner;

public class C {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        for (int ti=0; ti<T; ti++) {
            int n = in.nextInt();
            long k = in.nextLong();
            long[] a=new long[n];
            for (int i=0;i<n;i++) a[i]=in.nextLong();
            HashSet<Integer> pows = new HashSet<>();
            boolean yes=true;
            for (long x : a) {
                int p=0;
                while (x > 0) {//System.out.println(x + " " + pows);
                    if (x % k != 0) {
                        if (x%k != 1 || pows.contains(p)) {
                            yes=false;break;
                        }
                        pows.add(p);
                    }
                    x = x / k;
                    p++;
                }
                if (!yes) break;
            }
            System.out.println(yes? "YES": "NO");
        }
    }
}
