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.Collections;
import java.util.HashSet;
import java.util.Scanner;

public final class Third {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while(t -- > 0) {
            int n = in.nextInt();
            int k = in.nextInt();
            ArrayList<Long> arr = new ArrayList<>();
            for(int i = 0; i < n; i++) {
                arr.add(in.nextLong());
            }
            Collections.sort(arr, (a, b) -> {
                return (int)(b - a);
            });
            int currPow = 0;
            HashSet<Integer> set = new HashSet<>();
            while(1e16 > Math.pow(k, currPow)) {
                currPow ++;
            }
            String ans = "YES";
            while(currPow >= 0) {
                int c = 0;
                for(int i = 0; i < arr.size(); i++) {
                    if(arr.get(i) >= Math.pow(k, currPow)) {
                        c ++;
                        arr.set(i, arr.get(i) - (long)Math.pow(k, currPow));
                    }
                }
                if(c > 1) {
                    ans = "NO";
                    break;
                } else {
                    currPow --;
                }
            }
            for(long a : arr) {
                if(a != 0) {
                    ans = "NO";
                    break;
                }
            }
            System.out.println(ans);
        }
    }
}
