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.IOException;
import java.io.InputStreamReader;
import java.lang.ref.ReferenceQueue;
import java.lang.ref.SoftReference;
import java.lang.reflect.InvocationTargetException;
import java.time.Instant;
import java.time.LocalDate;
import java.time.Period;
import java.util.*;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.ReentrantLock;

// * Definition for a binary tree node.
public class Main {
    public static final long MOD = 99997867;
    private String[] strs;
    public int gcd(int a, int b) {
        while (b != 0) {
            int temp = a;
            a = b;
            b = temp % b;
        }
        return a;
    }

    long powMod(long a, long p) {
        long ans = 1;
        while (p != 0) {
            if ((p & 1) == 1) ans = (ans*a) % MOD;
            p >>= 1;
            a = (a*a) % MOD;
        }
        return ans%MOD;
    }
    public int[] fibonaci(){
        int[] fib=new int[100];
        fib[0]=0;
        fib[1]=1;
        for(int i=2;i<=100;i++) fib[i]=fib[i-1]+fib[i-2];
        return fib;
    }
    public int find(int[] parent,int node){
        int height=0;
        int pre=parent[node];
        while(pre!=-1){
            pre=parent[pre];
            height++;
        }
        return height;
    }
    private static ReferenceQueue<MyObject> softRefQueue=new ReferenceQueue<>();
    public static class MyObject{

        @Override
        public String toString() {
            return "i am object";
        }

        @Override
        protected void finalize() throws Throwable {
            super.finalize();
            System.out.println("finalize方法执行了");
        }
    }

    boolean[] vis=new boolean[64];
    int count=0;
    @SuppressWarnings(value = "duplicate")
    //dp[i][0]=
    public void solve() {
    Scanner in=new Scanner(System.in);
    int t=in.nextInt();
    while(t-->0){
        int n=in.nextInt();
        int k=in.nextInt();
        boolean legal=true;
        long[] num=new long[n];
        Arrays.fill(vis,false);
        for(int i=0;i<n;i++) num[i]=in.nextLong();
        Arrays.sort(num);
        for(int i=0;i<n;i++){
            if(num[i]==0) continue;
            if(!isLegal(num[i],k)) {
                legal=false;
                System.out.println("NO");
                break;
            }
        }

        if(legal)System.out.println("YES");
    }
    }
    private boolean isLegal(long d,int k){
        int step=0;
        while(true){
            while(d%k==0){
                d/=k;
                step++;
            }
            if(!vis[step]){
                d--;
                vis[step]=true;
            }
            if(d==0) return true;
            else if(d%k!=0){
                vis[step]=false;
                return false;
            }
        }
    }

    public static void main(String[] args) {
//        try {
            new Main().solve();

    }
}
