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 codeforces {
	public static PrintWriter out;
	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		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(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			}
			catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	// The next two void functions are shuffle functions on an array.
	// This way, quick sort encounters the worst case scenario with low probability.
	// A reason for doing this is if we don't a primitive data array to an object array.
	// For safety reasons, it would be preferable to use object arrays unless doing so expends a lot of time.
	public static void shuffleLongArray(long[] arr) {
		int n = arr.length;
		Random rnd = new Random();
        for (int i=0; i < n; ++i) {
        	long tmp = arr[i];
        	int randomPos = i + rnd.nextInt(n-i);
        	arr[i] = arr[randomPos];
        	arr[randomPos] = tmp;
        }
	}
	public static void shuffleIntArray(int[] arr) {
		int n = arr.length;
		Random rnd = new Random();
        for (int i=0; i < n; ++i) {
        	int tmp = arr[i];
        	int randomPos = i + rnd.nextInt(n-i);
        	arr[i] = arr[randomPos];
        	arr[randomPos] = tmp;
        }
	}
	public static long inf = 1000000007L;
	public static void main (String[] args) throws IOException {
    	MyScanner sc = new MyScanner();
    	out = new PrintWriter(new BufferedOutputStream(System.out));
    	// Write your solution here.
    	int t = sc.nextInt();
    	for (int asap=0; asap < t; ++asap) {
    		int n = sc.nextInt(); long k = sc.nextLong();
    		long[] target = new long[n];
    		for (int i=0; i < n; ++i) {
    			target[i] = sc.nextLong();
    		}
    		// max exponent 53, so 54 checks
    		int possible = 1;
    		String[] blog = new String[n];
    		for (int i=0; i < n; ++i) {
    			String q = "";
    			long w = target[i];
    			while (w > 0L) {
    				long y = w/k;
    				long r = w % k;
    				if (r > 1L) {
    					possible = 0; break;
    				}
    				q = q + Long.toString(r);
    				w = y;
    			}
    			if (possible == 0) break;
    			while (q.length() < 54) {
    				q = q + "0";
    			}
    			blog[i] = q;
    		}
    		if (possible == 0) {
    			out.println("NO"); continue;
    		}
    		for (int j=0; j < 54; ++j) {
    			int lap = 0;
    			for (int i=0; i < n; ++i) {
    				if (blog[i].charAt(j) == '1') ++lap;
    			}
    			if (lap > 1) {
    				possible = 0; break;
    			}
    		}
    		if (possible == 0) out.println("NO");
    		else out.println("YES");
    	}
    	out.close();
	}
	public static long exp(long z, long e) {
		if (e == 0L) return 1L;
		else if (e == 1L) return z;
		long y = z;
		String g = Long.toBinaryString(e);
		for (int i=1; i < g.length(); ++i) {
			y = (y*y) % inf;
			if (g.charAt(i)=='1') y = (y*z) % inf;
		}
		return y;
	}
	public static int gcd(int a, int b) {
		if (a > b) {
			int tmp = a; a = b; b = tmp;
		}
		if (b % a == 0) return a;
		else return gcd(b % a, a);
	}
}