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.
	
	// you can also use imports, for example:
	import java.util.*;
	import java.io.*;
	
	public class Solution {
		public static void main(String args[]) {
			Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
			int test_cases = in.nextInt();
			for (int q = 0; q < test_cases; q++) {
				int n = in.nextInt(), m = in.nextInt();
				long[] arr = new long[n];
				for (int x = 0; x < n; x++)
					arr[x] = in.nextLong();
	
				HashSet<Integer> set = new HashSet();
				boolean flag = true;
				for (int x = 0; x < n; x++) {
					if (!print_fun(arr[x], m, set)) {
						flag = false;
						break;
					}
	
				}
				if (flag)
					System.out.println("YES");
				else
					System.out.println("NO");
			}
		}
	
		private static boolean print_fun(long ele, int m, HashSet<Integer> set) {
			int i = 0;
			while (ele != 0) {
				
				while (ele % m == 0) {
					ele /= m;
					i++;
				}
	
				if (set.contains(i)) {
					return false;
				}
				set.add(i);
				ele -= 1;
			}
	
			return true;
		}
	
	}
