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.HashMap;

 

public class S1Q2 {

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

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int t = Integer.parseInt(br.readLine());

        String[] a1,ai;

        int n,x;

        for (int i = 0; i < t; i++) {

            a1 = br.readLine().split(" ");

            n = Integer.parseInt(a1[0]);

            x = Integer.parseInt(a1[1]);

            ai = br.readLine().split(" ");

            //调方法

            System.out.println(constructList(n,x,ai));

        }

 

 

    }

    public static String constructList(int n,int x,String[] ai){

        HashMap<Integer, Integer> dic = new HashMap<>();

        for (int i = 0; i < n; i++) {

            Long a = Long.parseLong(ai[i]);

            //boolean flag = true;

            if(a==1){

                if(dic.containsKey(0)){

                    return "NO";

                }else{

                    dic.put(0,1);

                }

            }else if (a > 1){

                int m=0;

                while (a>=1){

                    if(a%x == 1){

                        if(dic.containsKey(m)){

                            return "NO";

                        }else{

                            dic.put(m,1);

                        }

                        a=a-1;

 

                    }else if(a%x == 0){

                        if (a == x){

                            a = 0L;

                        }else{

                            a=a/x;

                        }

                        m++;

                    }else{

                        return  "NO";

                    }

                }

                if(dic.containsKey(m)){

                    return "NO";

                }else{

                    dic.put(m,1);

                }

            }

        }

        return "YES";

    }

}
		   	 		  	  	   	 	   	 		  	