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 Main {
    static PrintWriter out = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    List<Integer>[] graph;
    public static void main(String[] args) throws IOException {
        Main main = new Main();
        int t = sc.nextInt();
        for(int i=0;i<t;i++){
            String res = main.solve()>=0?"YES":"NO";
            out.println(res);
        }
//        main.solve();
        out.flush();
//        System.out.println(main.solve());
    }
    long solve() throws IOException{
        int n = sc.nextInt(); long k = sc.nextLong();
        long[] a = new long[n];
        for(int i=0;i<n;i++) a[i]= sc.nextLong();
        boolean[] power = new boolean[100];
        for(long w:a){
            if(w==0) continue;
            List<int[]> pos = helper(w,k);
            for(int[] res:pos){
                if(res[0]>1) return -1;
                if(res[0]==1){
                    int p = res[1];
                    if(power[p]) return -1;
                    power[p] = true;
                }
            }
        }
        return  1;
    }
    List<int[]> helper(long a, long k){
        List<int[]> ans = new LinkedList<>();
        long base = 1;
        int p = 0;
        while(base<=a) {
            base *= k; p++;
        }
        p--;
        base /= k;
        while(a>0){
            long cur = a/base;
            ans.add(new int[]{(int)cur,p});
            p--;
            a = a%base;
            base /= k;
        }
        return ans;
    }
    int[] getLine() throws IOException{
        String[] buf = reader.readLine().split(" ");
        int[] ans = new int[buf.length];
        for(int i=0;i<ans.length;i++) ans[i] = Integer.parseInt(buf[i]);
        return ans;
    }
}

