Problem: Ashish has two strings a and b, each of length n, and an integer k. The strings only contain lowercase English letters.

He wants to convert string a into string b by performing some (possibly zero) operations on a.

In one move, he can either 

  * choose an index i (1 ≤ i≤ n-1) and swap a_i and a_{i+1}, or 
  * choose an index i (1 ≤ i ≤ n-k+1) and if a_i, a_{i+1}, …, a_{i+k-1} are all equal to some character c (c ≠ 'z'), replace each one with the next character (c+1), that is, 'a' is replaced by 'b', 'b' is replaced by 'c' and so on. 



Note that he can perform any number of operations, and the operations can only be performed on string a. 

Help Ashish determine if it is possible to convert string a into b after performing some (possibly zero) operations on it.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. The description of each test case is as follows.

The first line of each test case contains two integers n (2 ≤ n ≤ 10^6) and k (1 ≤ k ≤ n).

The second line of each test case contains the string a of length n consisting of lowercase English letters.

The third line of each test case contains the string b of length n consisting of lowercase English letters.

It is guaranteed that the sum of values n among all test cases does not exceed 10^6.

Output

For each test case, print "Yes" if Ashish can convert a into b after some moves, else print "No".

You may print the letters of the answer in any case (upper or lower).

Example

Input


4
3 3
abc
bcd
4 2
abba
azza
2 1
zz
aa
6 2
aaabba
ddddcc


Output


No
Yes
No
Yes

Note

In the first test case it can be shown that it is impossible to convert a into b.

In the second test case,

"abba" \xrightarrow{inc} "acca" \xrightarrow{inc} … \xrightarrow{inc} "azza".

Here "swap" denotes an operation of the first type, and "inc" denotes an operation of the second type.

In the fourth test case,

"aaabba" \xrightarrow{swap} "aaabab" \xrightarrow{swap} "aaaabb" \xrightarrow{inc} … \xrightarrow{inc} "ddaabb" \xrightarrow{inc} … \xrightarrow{inc} "ddddbb" \xrightarrow{inc} … \xrightarrow{inc} "ddddcc".

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
public class prob4
{
    static PrintWriter out=new PrintWriter((System.out));
    public static void main(String args[])throws IOException
    {
        Reader sc=new Reader();
        int t=sc.nextInt();
        while(t-- >0){
            int n=sc.nextInt(),k=sc.nextInt();     
            char[] a=sc.next().toCharArray();
            char[] b=sc.next().toCharArray();
            int[] af=new int[27];
            int[] bf=new int[27];
            for(int i=0;i<n;i++){
                af[a[i]-'a']++;
                bf[b[i]-'a']++;
            }
            boolean ok=true;
            for(int i=0;i<26;i++){
                if(af[i]<bf[i] || (af[i]-bf[i])%k!=0){
                    ok=false;
                    break;
                }
                af[i+1]+=(af[i]-bf[i]);
            }

            
            if(ok) out.println("Yes");
            else out.println("No");
        }
    	out.close();
    }
    // lis => nlogn
    private static int lis(int[] arr,int l,int r){
        if(l==r) return 1;
        int[] len=new int[r-l+1];
        int k=0;
        len[0]=arr[l];
        for(int i=l+1;i<=r;i++){
            if(arr[i]>len[k]){
                k++;
                len[k]=arr[i];
            }else{
                bin(len,k,arr[i]);
            }
        }
        return k+1;
    }

    private static void bin(int[] len,int max,int val){
        int s=0,e=max,mid,idx=-1;
        while(s<=e){
            mid=s+(e-s)/2;
            if(len[mid]>val){
                idx=mid;
                e=mid-1;
            }else if(len[mid]<val){
                s=mid+1;
            }else{
                return;
            }
        }
        len[idx]=val;
    }
    
    // To compute x^y under modulo m 
    static int power(int x, int y, int m)  
    { 
        if (y == 0) 
            return 1; 
          
        int p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    } 
    
    
    /*
    int n=sc.nextInt();
        	char[] s=sc.next().toCharArray();
        	int res=0;
        	for(int i=1;i<n;i++){
        		if( (s[i]=='1') && s[i]==s[i-1]) res++;
        	}
        	out.println(res);
    */
    
 
    public static long pow(long a,long b){
        if(b==0){
            return 1;
        }
        if(b%2==0){
            return pow(a*a,b/2);
        }else{
            return a*pow(a*a,b/2);
        }
    }
    static class Reader 
    { 
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        public String next()
        {
            while(!st.hasMoreTokens())
            {
                try
                {
                    st=new StringTokenizer(br.readLine());
                }
                catch(Exception e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        public int nextInt()
        {
            return Integer.parseInt(next());
        }
        public long nextLong()
        {
            return Long.parseLong(next());
        }
        public double nextDouble()
        {
            return Double.parseDouble(next());
        }
        public String nextLine()
        {
            try
            {
                return br.readLine();
            }
            catch(Exception e)
            {
                e.printStackTrace();
            }
            return null;
        }
        public boolean hasNext()
        {
            String next=null;
            try
            {
                next=br.readLine();
            }
            catch(Exception e)
            {
            }
            if(next==null)
            {
                return false;
            }
            st=new StringTokenizer(next);
            return true;
        }
    } 
}