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 TEST
{
    static class Reader
    {
        final private int BUFFER_SIZE = 1 << 16;
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
        public Reader() 
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
        public Reader(String file_name) throws IOException 
        {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
        public String nextString() throws IOException
        {
            String str00 = scan.next();
            return str00;
        }
        public String readLine() throws IOException
        {
            byte[] buf = new byte[64];
            int cnt = 0, c;
            while ((c = read()) != -1)  
            {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }
        String next() throws IOException 
        {
            int c;
            for (c = read(); c <= 32; c = read());
            StringBuilder sb = new StringBuilder();
            for (; c > 32; c = read())
            {
                sb.append((char) c);
            }
            return sb.toString();
        }
        public int nextInt() throws IOException 
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do 
            {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        } 
        public long nextLong() throws IOException 
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');           
            if (c == '.')
            {
                while ((c = read()) >= '0' && c <= '9')
                {
                    ret += (c - '0') / (div *= 10);
                }
            } 
            if (neg)
                return -ret;
            return ret;
        }
        private void fillBuffer() throws IOException 
        {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
        private byte read() throws IOException 
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        } 
        public void close() throws IOException 
        {
            if (din == null)
                return;
            din.close();
        }
        public int[] nextArray(int n) throws IOException 
        {
            int[] a=new int[n];
            for (int i=0;i<n;i++)
            {
                a[i]=nextInt();
            }
            return a;
        }
        public long[] nextArrayl(int n) throws IOException 
        {
            long[] a = new long[n];
            for (int i=0;i<n;i++)
            {
                a[i]=nextLong();
            }
            return a;
        }
    }
    public static void main(String args[])throws IOException
    {
        Reader re=new Reader();
        int t=re.nextInt();
        while(t-->0)
        {
            int n=re.nextInt();
            int k=re.nextInt();
            char a[]=re.next().toCharArray();
            char b[]=re.next().toCharArray();
            int c1[]=new int[26];
            int c2[]=new int[26];
            int c=0,flag=0;
            for(int i=0;i<n;i++)
            {
                c1[a[i]-'a']++;
                c2[b[i]-'a']++;
            }
            for(int i=0;i<26;i++)
            {
                if(c1[i]==c2[i])
                {
                    continue;
                }
                if(c2[i]>(c1[i]+c))
                {
                    flag=1;
                    break;
                }
                else if((c1[i]-c2[i])%k==0 && (c>(c1[i]-c2[i])))
                {
                    c=c-(c2[i]-c1[i]);
                }
                else
                {
                    if((c1[i]-c2[i])%k!=0)
                    {
                        flag=1;
                        break;
                    }
                    else
                    {
                        c+=(c1[i]-c2[i]);
                    }
                }
            }
            if(flag==1 || c!=0)
            {
                System.out.println("No");
            }
            else
            {
                System.out.println("Yes");
            }
        }
    }
}