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.*;
import java.util.*;

public class C {
	public static int search(List<Long> arr, long k, HashSet<Integer> hs) {
		int ind=0, ans=-1, n=arr.size();
		while(ind < n) {
			if(hs.contains(ind)) ind++;
			else if((k-arr.get(ind))>=0) {
				ans = ind;
				ind++;
			}
			else break;
		}
		return ans;
	}
	
	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		while(t-- >0) {
			String[] str;
			str = br.readLine().split(" ");
			int n = Integer.parseInt(str[0]), k = Integer.parseInt(str[1]);
			long sum=0, max;
			Long[] a = new Long[n];
			HashSet<Integer> hs = new HashSet<>();
			str = br.readLine().split(" ");
			for(int i=0; i<n; i++) {
				a[i] = Long.parseLong(str[i]);
				sum += a[i];
			}
			Arrays.sort(a);
			max = a[n-1];
			List<Long> l = new ArrayList<>();
			l.add(1L);
			while(l.get(l.size()-1)<=max) l.add(l.get(l.size()-1)*k);
			boolean possible = true;
			if(sum!=0) {
				for(int i=0; i<n&&possible; i++) {
					long cur = a[i];
					while(cur>0 && possible) {
						int ind = search(l, cur, hs);
						if(ind == -1) possible = false;
						else {
							cur -= l.get(ind);
							hs.add(ind);
						}
					}
					if(cur!=0) possible = false;
				}
			}
			if(possible) System.out.println("YES");
			else System.out.println("NO");
		}
	}
}
