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 noob {
    InputReader in;
    final long mod=1000000007;
    long m=998244353;
    StringBuilder sb;
    public static void main(String[] args) throws java.lang.Exception {
        new noob().run();
    }
    void run() throws Exception {
        in=new InputReader(System.in);
        sb = new  StringBuilder();
        int t=i();
        for(int i=1;i<=t;i++)
            solve();
        System.out.print(sb);
    } 
    void solve() {
        int i, j;
        int n=i(), k=i(),ans=0;
        String a=s(), b=s();
        char s1[]=new char[n];
        char s2[]=new char[n];
        for(i=0;i<n;i++)
            s1[i]=a.charAt(i);
        for(i=0;i<n;i++)
            s2[i]=b.charAt(i);
        Arrays.sort(s1);
        Arrays.sort(s2);
        int freq1[]=new int[26];
        int freq2[]=new int[26];
        for(i=0;i<n;i++)
            freq1[a.charAt(i)-'a']++;
        for(i=0;i<n;i++)
            freq2[b.charAt(i)-'a']++;
        int spare=0;
        for(i=0;i<26;i++) {
            if(freq1[i]==freq2[i])
                continue;
            else if(freq1[i]>freq2[i]) {
                spare+=freq1[i]-freq2[i];
                if(spare%k != 0) {
                    sb.append("No\n");
                    return;
                }
            }
            else {
                int req=freq2[i]-freq1[i];
                //System.out.println(spare+"  "+req+"  "+i);
                if(req%k!=0 || spare<req) {
                    sb.append("No\n");
                    return;
                }
                spare-=req;
            }
            //System.out.println(spare+"  "+i);
        }
        ans=1;
        if(ans==0)
            sb.append("No\n");
        else
            sb.append("Yes\n");
    }
    long power(long x, long y, long p) {  
        long res = 1;       
        x = x % p;  
        if (x == 0) return 0;   
        while (y > 0) {
            if((y & 1)==1) 
                res = (res * x) % p;  
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res%p; 
    } 
    int gcd(int a, int b) {
        return (b==0)?a:gcd(b,a%b);
    }
    String s(){return in.next();}
    int i(){return in.nextInt();}
    long l(){return in.nextLong();}
    double d(){return in.nextDouble();}
    class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        public InputReader(InputStream stream) {
            this.stream = stream;
        }
        public int read() {
            if (numChars == -1)
                throw new UnknownError();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new UnknownError();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }
        public int peek() {
            if (numChars == -1)
                return -1;
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    return -1;
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar];
        }
        public void skip(int x) {
            while (x-->0)
                read();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public String nextString() {
            return next();
        }
        public String next() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuffer res = new StringBuffer();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
 
            return res.toString();
        }
        public String nextLine() {
            StringBuffer buf = new StringBuffer();
            int c = read();
            while (c != '\n' && c != -1) {
                if (c != '\r')
                    buf.appendCodePoint(c);
                c = read();
            }
            return buf.toString();
        }
        public double nextDouble() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            double res = 0;
            while (!isSpaceChar(c) && c != '.') {
                if (c == 'e' || c == 'E')
                    return res * Math.pow(10, nextInt());
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            if (c == '.') {
                c = read();
                double m = 1;
                while (!isSpaceChar(c)) {
                    if (c == 'e' || c == 'E')
                        return res * Math.pow(10, nextInt());
                    if (c < '0' || c > '9')
                        throw new InputMismatchException();
                    m /= 10;
                    res += (c - '0') * m;
                    c = read();
                }
            }
            return res * sgn;
        }
        public boolean hasNext() {
            int value;
            while (isSpaceChar(value = peek()) && value != -1)
                read();
            return value != -1;
        }
        private boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }
    }
}