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.HashSet;
import java.util.Scanner;
import java.util.Set;

public class conArray {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        sc.nextLine(); 
        String[] strings = new String[2 * t];
        for (int i = 0; i < 2 * t; i++) {
            strings[i] = sc.nextLine();
        }
        for (int i = 0; i < 2 * t; i++) {
            int n = 0;
            int x = 0;
            if (i % 2 == 0) {
                String[] s = strings[i].split(" ");
                n = Integer.parseInt(s[0]);
                x = Integer.parseInt(s[1]);
            }
            if (n != 0 && x != 0) {
                String[] s = strings[i + 1].split(" ");
                long[] longs = new long[n];
                for (int j = 0; j < n; j++) {
                    longs[j] = Long.parseLong(s[j]);
                }
                boolean flag = true;
                int id = 0;
                Set<Integer> set = new HashSet<Integer>();
                for (int j = 0; j < n && flag; j++) {
                    int index = 0;
                    if (longs[j] == 0) {
                        continue;
                    } else if (longs[j] == 1) {
                        flag = set.add(index);
                        continue;
                    }
                    long tmp = longs[j];
                    while (flag && longs[j] > 1 && tmp > 0) {
                        index = 0;
                        if (tmp == 1) {
                            flag = set.add(index);
                            break;
                        }
                        boolean flag2 = true;
                        long a = tmp;
                        while (a > 0 && flag2) {
                            if ((a % x) == 0 || (a % x) == 1) {
                                a = a / x;
                                index++;
                            } else {
                                flag = false;
                                break;
                            }
                        }
                        flag = set.add(index - 1);
                        tmp = tmp - (long) Math.pow(x, (index - 1));
                    }
                }
                if (flag) {
                    System.out.println("YES");
                } else {
                    System.out.println("NO");
                }
            }
        }
    }
}

 	        	   	   			 	 					