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.*;
import java.io.*;
public class Main
{
    public static void main(String []args) throws IOException {
    	FastScanner in  = new FastScanner(System.in);
    	PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
    	solve(in, out);
    	in.close();
    	out.close();
    }
    static long gcd(long a, long b){ return (b==0) ? a : gcd(b, a%b); }
    static int gcd(int a, int b){ return (b==0) ? a : gcd(b, a%b); }
    private static int fact(int n) { int ans=1; for(int i=2;i<=n;i++) ans*=i; return ans; }
    static class FastScanner{
	BufferedReader reader;
	StringTokenizer st;
	FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}
	String next(){while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}
	     st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
	String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}
	int    nextInt()   {return Integer.parseInt(next());}
	long   nextLong()  {return Long.parseLong(next());}
	double nextDouble(){return Double.parseDouble(next());}
	char   nextChar()  {return next().charAt(0);}
	int[]  nextIntArray(int n)         {int[] arr= new int[n];   int i=0;while(i<n){arr[i++]=nextInt();}  return arr;}
	long[] nextLongArray(int n)        {long[]arr= new long[n];  int i=0;while(i<n){arr[i++]=nextLong();} return arr;}
	int[]  nextIntArrayOneBased(int n) {int[] arr= new int[n+1]; int i=1;while(i<=n){arr[i++]=nextInt();} return arr;}
	long[] nextLongArrayOneBased(int n){long[]arr= new long[n+1];int i=1;while(i<=n){arr[i++]=nextLong();}return arr;}
	void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}
    }
    
    /********* SOLUTION STARTS HERE ************/
    
    private static void solve(FastScanner in, PrintWriter out){
        int t = in.nextInt();
        while(t-->0){
            int n = in.nextInt();
            int k = in.nextInt();
            long a[] = new long[n];
            for(int i=0;i<n;i++) a[i] = in.nextLong();
            int f=0;
            boolean res = true;
            boolean vis[] = new boolean[60];
            while(f==0){
                f=1;
                for(int i=0;i<n;i++){
                    int tmp = biggest_power_k(a[i],k);
                    // System.out.println("ans: "+tmp);
                    if(tmp!=-1 && vis[tmp]) {res = false;break;}
                    else if(tmp!=-1) vis[tmp] = true;
                    // long tmp2 = 
                    a[i]-=(long)Math.pow(k, tmp);
                    if(a[i]>0) f=0;
                    // System.out.println("ans: "+tmp+" "+a[i]+" "+Math.pow((long)k, tmp));
                }
                // for(int )
            }
            out.println(res?"YES":"NO");
        }
    }
    static int biggest_power_k(long n,int k){
        int i=0;
        long tmp = power(k,i);
        // System.out.println("tmp: "+tmp+" "+n);
        while(tmp<=n){
            ++i;
            tmp = power(k,i);
            // System.out.println("tmp: "+tmp+" "+i);
        }
        return i-1;
    }
    static long power(long k,int n){
        long ans=1L;
        while(n>0){
            if(n%2==1) ans*=k;
            k*=k;
            n>>=1;
        }
        return ans;
    }
    /************* SOLUTION ENDS HERE **********/
}