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.util.ArrayList;
import java.util.Scanner;

public class practice {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();//测试数据组数
        ArrayList<String> result = new ArrayList<>();//结果值
        for (int i = 0 ; i < t ; i++) {
            //接收数据n,x
            int n = sc.nextInt();
            int x = sc.nextInt();
            ArrayList<Long> expect =new ArrayList<>();
            ArrayList<Integer> powers =new ArrayList<>();//放幂次方数
            for( int j =0 ; j < n ; j++){
                //接收预期数组
                expect.add(sc.nextLong());
            }
            int success = 0;
            for(Long it: expect){
                if(it == 0){
                    success++;
                    continue;
                }else if( it ==1 && !powers.contains(0)){
                    success++;
                    powers.add(0);
                    continue;
                }else if(it < x){
                    continue;
                } else{
                    boolean b = surplus(it,x,powers);
                    if(b == true){
                        success++;
                    }
                }
            }
            if(success == expect.size()){
                result.add("YES");
            }else{
                result.add("NO");
            }

        }
        for (String it:result) {
            System.out.println(it);
        }
    }

    public static boolean surplus(Long x,int n, ArrayList<Integer> powers){
        Long r = x;//商
        int num = 0;//计数
        Long y =x;//被除数
        Long before = 0L;
        for( Long s = 1L; (s != 0 || r !=  1 ) || before != x; ){
            if(y<0){
                return false;
            }
            if(y == 0){
                break;
            }
            if(y == 1 && !powers.contains(0)){
                powers.add(0);
                break;
            }else if(y == 1 && powers.contains(0)){
                return false;
            }
            //没有被整除到最后时
            r = y/n;
            s = y%n;
            y=r;
            num++;
            if(r<n){
                //本次结束
                Long pow = new Double(Math.pow(n,num)).longValue();
                if(y == pow && y != n){
                    //若同样幂次，则返回失败
                    return false;
                }
                before = before+pow;
                y= x - before;
                if(!powers.contains(num)){
                    //该幂次未用过
                    powers.add(num);
                }else{
                    //有重复的幂次，返回失败
                    return false;
                }
                num = 0;
            }else if(s!= 0 && r < n){
                //没办法被整除
                return false;
            }
        }
        return true;
    }
}
   		     				  	 	 	 	 		    	