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 cf1312_EduDiv2_C {
	
	public static String convert(long val, int base) {
		if (val == 0)
			return "";
		if (val % base > 1)
			return "N";
		return convert(val / base, base) + "" + (val % base); 
	}
	
	public static boolean solve(int n, int k, long[] arr) {
	    	
    	boolean[] cnt = new boolean[60];
    	for (int i = 0; i < n ; i++) {
    		if (arr[i] == 0) continue;
    		StringBuilder s = new StringBuilder(convert(arr[i], k));
    		char[] str = s.reverse().toString().toCharArray();
    		// System.out.println(Arrays.toString(str) + " " + arr[i]);
       		for (int j = 0; j < str.length; j++) {
    			if (str[j] == 'N') return false;
       			if (str[j] == '1') {
    				// System.out.println(j);
    				if (cnt[j]) return false;
    				cnt[j] = true;
    				
    			}
    		}
    	}
    	return true;
	}
	public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        // 100000 
        // 001100 
        // 101100 = 
        int t = in.nextInt();
        for ( ; t > 0; t--) {
        	int n = in.nextInt();
        	int k = in.nextInt();
        	long[] arr = new long[n];
        	for (int i = 0; i < n; i++) {
        		arr[i] = in.nextLong();
        	}
        	
        	if (solve(n, k, arr))
        		out.println("Yes");
        	else
        		out.println("No");
        	
        	
        }
        out.close();
    }
    
    public static void shuffle(int[] arr) {
    	Random rgen = new Random();
    	
    	for (int i = 0; i < arr.length; i++) {
    		int rPos = rgen.nextInt(arr.length);
    		int temp = arr[i];
    		arr[i] = arr[rPos];
    		arr[rPos]=temp;
    	}
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String nextToken() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine() {
            st = null;
            try {
                return br.readLine();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            return null;
        }
        
        String next() {
    		while (st == null || !st.hasMoreElements()) {
    			try {
    				st = new StringTokenizer(br.readLine());
    			} catch (IOException e) {
    				e.printStackTrace();
    			}
    		}
    		return st.nextToken();
    	}

        int nextInt() {
            return Integer.parseInt(nextToken());
        }

        long nextLong() {
            return Long.parseLong(nextToken());
        }

        double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
