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 void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int tst = Integer.parseInt(br.readLine());
        while (tst-->0){
            String[] str = br.readLine().split(" ");
            int n = Integer.parseInt(str[0]), k = Integer.parseInt(str[1]);
            long[] arr = new long[n];
            StringTokenizer st = new StringTokenizer(br.readLine());
            for(int i = 0; i<n; i++){
                arr[i] = Long.parseLong(st.nextToken());
            }
            int[] freq_pow = new int[60];
            boolean poss = true;
            outer:for(int i = 0; i<n; i++){
                long z = arr[i];
                int cnt = 0;
                while(z>0){
                    if(z%k!=0){
                        freq_pow[cnt]++;
                        z--;
                        if(z%k!=0){
                            poss = false;
                            break outer;
                        }
                    }
                    z/=k;
                    cnt++;
                }
            }
            if(poss)
            for(int i = 0; i<60; i++){
                if(freq_pow[i]>=2){
                    poss = false;
                    break;
                }
            }
            if(poss) sb.append("YES");
            else sb.append("NO");
            sb.append('\n');
        }
        System.out.println(sb);
    }
}