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.
// package codeforce.edu83;

import java.io.PrintWriter;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Scanner;

public class C {
    //    MUST SEE BEFORE SUBMISSION
//    check whether int part would overflow or not, especially when it is a * b!!!!
//    check if top down dp would cause overflow or not
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
//        int t = 1;
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }

    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt(), k = in.nextInt();
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextLong();
        }
        long max = 0;
        LinkedList<Long> q = new LinkedList<>();
        long[] ok = new long[64];
        for (int i = 0; i < n; i++) {
            max = Math.max(max, arr[i]);
        }
        long mut = 1;
        ok[0] = 1;
        for (int i = 1;; i++) {
            ok[i] = ok[i - 1] * k;
            if (ok[i] > max) break;
        }
        Arrays.sort(arr);
        for(int i = n - 1; i >= 0; i--){
            long mis = arr[i];
            for (int j = 63; j >= 0; j--) {
                if (ok[j] == 0 || ok[j] > mis) continue;
                mis -= ok[j];
                ok[j] = 0;
            }
            if (mis > 0){
                out.println("NO");
                return;
            }
        }
        out.println("YES");
    }
}
