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.
/* The idea is that decompose every number in base k. If for any number remainder is greater than 1 or remainder occurs with same 
powers in 2 numbers then ans is NO because we can add a power of exactly once.

*/
import java.util.*;
import java.io.*;
public class lp{ 
static PrintWriter out = new PrintWriter(System.out);
static boolean check(long num,int k,int ct[])
{

        int c=0;
        while(num>0)
         {
             int x = (int)(num%k);
             if(x>1)
                return false;
             else if(x==1)
               {
                  if(ct[c]==1)
                      return false;
                  ct[c]=1;
               }
             c++;
             num = num/k;
         }
      return true;
}
public static void main(String[] args){

int t = ni();


for(int o=1;o<=t;o++){

int n = ni();
int k = ni();

long a[] = al(n);
int ct[] = new int[100];
int f=1;

for(int i=0;i<n;i++)
{
         boolean z = check(a[i],k,ct);
         if(z==false)
           { f=0; break;}
}
   if(f==1)
     pn("YES");
   else
     pn("NO");
}
   out.flush();
} 
static int[] ai(int n)  // it will give in array of size n
{
    int a[] = new int[n];
    for(int i=0;i<n;i++)
        a[i] = ni();
    return a;
}
static long[] al(int n)  // it will give in array of size n
{
    long a[] = new long[n];
    for(int i=0;i<n;i++)
        a[i] = nl();
    return a;
}

static void p(Object o)
{
   out.print(o);
}
static void pn(Object o)
{
    out.println(o);
}
static int abs(int x)
{
    return x>0 ? x : -x;
}
static long gcd(long a,long b)
{
       if(b%a==0)
          return a;
          return gcd(b%a,a);
}
static void subtract_1(char s[]) // it will subtract 1 from the given number.  number should be positive
{
      if(s[0]=='0') // number is zero
          return;
     int n = s.length,i=n-1;
     while(s[i]=='0')
       i--;
     s[i] = (char)((int)(s[i]-'0') + 47);
   
     for(int j=i+1;j<n;j++)
         s[j]='9';
}
static long pow(long a,long b,long md)
{ 
    long ans=1;
    while(b>0)
    {
         if(b%2==1)
           ans = (ans*a)%md;
           a = (a*a)%md;
           b = b/2;
    }
   return ans;
}
static long min(long a,long b){
 
     return a<b ? a : b;
}
static long max(long a,long b){
     return a>b ? a : b;
}
static boolean pal(String s)
{
   int n = s.length(),i1=0,i2=n-1;
 
     while(i1<i2)
     {
        if(s.charAt(i1)!=s.charAt(i2))
             return false;
             i1++; i2--;
     }
       return true;
}
static String rev(String r)
{
    String s = "";
    int i= r.length()-1;
 
    while(i>=0)
    {
        s=s+r.charAt(i);
        i--;   
    }
       return s;
}
static FastReader sc=new FastReader(); 
 
      static int ni(){
                 int x = sc.nextInt();
                 return(x);
        }
      static long nl(){
              long x = sc.nextLong();
              return(x);
         }
      static String n(){
                 String str = sc.next();
                     return(str);
       }
     static String ns(){
                 String str = sc.nextLine();
                   return(str);
      }
     static double nd(){
               double d = sc.nextDouble();
                 return(d);
       }
  static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            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; 
        } 
    } 
}