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.math.BigInteger;
import java.util.*;
import java.io.*;
import java.text.*;

public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 32768);
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer t;

    static String sn() {
        while (t == null || !t.hasMoreTokens()) {
            try {
                t = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }
        return t.nextToken();
    }

    static int ni() {
        return Integer.parseInt(sn());
    }

    static long nlo() {
        return Long.parseLong(sn());
    }

    public static void main(String args[]) {
        int t = ni();
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    static long mod = (long) 998244353;

    static long pro(long a, long b) {
        return (a % mod * b % mod) % mod;
    }

    static long add(long a, long b) {
        return (a + b) % mod;
    }

    static long sub(long a, long b) {
        return ((a % mod - b % mod) + mod) % mod;
    }

    static int maxn = 1000005;

    static int pow(int n) {
        return (int) (Math.log10(n) / Math.log10(2));
    }

    static long modpow(long x, long y) {
        long res = 1l;
//        x %= mod;
        while (y > 0) {
            if (y % 2 != 0)
                res *= x;
            x = x * x;
            y /= 2;
        }
        return res;
    }

    static class pair {
        String a;
        int b;

        pair(String x, int y) {
            a = x;
            b = y;
        }
    }

    static long modInverse(long n) {
        BigInteger n1 = new BigInteger(Long.toString(n));
        String s = (n1.modInverse(new BigInteger(Long.toString(mod)))).toString();
        return Long.parseLong(s);
    }

    static boolean v[] = new boolean[1000005];

    static void seive() {
        v[1] = true;
        for (int i = 2; i < 1000005; i++) {
            if (!v[i]) {
                for (int j = 2 * i; j < 1000005; j += i)
                    v[j] = true;
            }
        }
    }

    static boolean isprime(long n) {
        if (n == 1)
            return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static DecimalFormat df = new DecimalFormat("0.000000000000000000000");

    static void solve() {
        int n = ni();
        int k = ni();
        long a[] = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = nlo();
        }
        long max = modpow(10, 16);
        long p = k;
        while (p <= max)
            p *= k;
        while (p != 0) {
            int cnt = 0;
            for (int i = 0; i < n; i++) {
                if (a[i] >= p) {
                    a[i] -= p;
                    cnt++;
                }
            }
            if (cnt > 1) {
                out.println("NO");
                return;
            }
            p /= k;
        }
        for (int i = 0; i < n; i++) {
            if (a[i] != 0) {
                out.println("NO");
                return;
            }
        }
        out.println("YES");
    }
}

