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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;

public class AddingPowers {

	public static void main(String[] args) throws IOException {

		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		String[] input;
		int t = Integer.parseInt(reader.readLine());
		int n, k;
		long[] arr;
		StringBuilder builder = new StringBuilder();
		while (t-- > 0) {
			input = reader.readLine().split(" ");
			n = Integer.parseInt(input[0]);
			k = Integer.parseInt(input[1]);
			input = reader.readLine().split(" ");
			arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = Long.parseLong(input[i]);
			}
			builder.append(isSequencePossible(arr, k) ? "YES" : "NO");
			builder.append("\n");
		}
		System.out.println(builder.toString());
	}

	private static boolean isSequencePossible(long[] arr, int k) {
		Set<Long> set = new HashSet<>();
		List<Long> sequence;
		Map<Long, List<Long>> sequenceMap = new HashMap<>();
		for (int i = 0; i < arr.length; i++) {
			sequence = findPowerSequence(arr[i], k , sequenceMap);
			for (long s : sequence) {
				if (set.contains(s)) {
					return false;
				}
				set.add(s);
			}
		}
		return true;
	}

	private static List<Long> findPowerSequence(long n, long k, Map<Long, List<Long>> sequenceMap) {

		if (sequenceMap.containsKey(n)) {
			return sequenceMap.get(n);
		}

		List<Long> sequence = new ArrayList<>();

		if (n != 0) {
			if (n == 1) {
				sequence.add(0l);
			} else {
				long m = n;
				if (m % k != 0) {
					sequence.add(0l);
					m = n - 1;
				}
				if (m % k != 0) {
					sequence.add(0l);
				} else {
					List<Long> subSequence = findPowerSequence(m / k, k, sequenceMap);
					for (int i = 0; i < subSequence.size(); i++) {
						sequence.add(subSequence.get(i) + 1L);
					}
				}
			}
		}
		sequenceMap.put(n, sequence);
		return sequence;
	}
}