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.util.*;
import java.io.*;

public class CFC {
  BufferedReader br;
  PrintWriter out;
  StringTokenizer st;
  boolean eof;
  private static final long MOD = 1000L * 1000L * 1000L + 7;
  private static final int[] dx = {0, -1, 0, 1};
  private static final int[] dy = {1, 0, -1, 0};
  private static final String yes = "Yes";
  private static final String no = "No";

  void solve() {
    int T = nextInt();
//    int T = 1;
    for (int i = 0; i < T; i++) {
      helper();
    }
  }

  void helper() {
    int n = nextInt();
    int k = nextInt();
    char[] a = nextString().toCharArray();
    char[] b = nextString().toCharArray();
    int[] h1 = findHist(a);
    int[] h2 = findHist(b);
    int prev = 0;
    for (int i = 0; i < h1.length; i++) {
      int sum = prev + h1[i];
      if (sum < h2[i]) {
        outln(no.toUpperCase());
        return;
      }
      int diff = sum - h2[i];
      if (diff % k != 0) {
        outln(no.toUpperCase());
        return;
      }
      prev = sum - h2[i];
    }

    outln(yes.toUpperCase());
  }

  int[] findHist(char[] arr) {
    int[] res = new int[26];
    for (char c : arr) {
      res[c - 'a']++;
    }
    return res;
  }

  void shuffle(int[] a) {
    int n = a.length;
    for(int i = 0; i < n; i++) {
      int r = i + (int) (Math.random() * (n - i));
      int tmp = a[i];
      a[i] = a[r];
      a[r] = tmp;
    }
  }
  long gcd(long a, long b) {
    while(a != 0 && b != 0) {
      long c = b;
      b = a % b;
      a = c;
    }
    return a + b;
  }
  private void outln(Object o) {
    out.println(o);
  }
  private void out(Object o) {
    out.print(o);
  }
  private void formatPrint(double val) {
    outln(String.format("%.9f%n", val));
  }
  public CFC() {
    br = new BufferedReader(new InputStreamReader(System.in));
    out = new PrintWriter(System.out);
    solve();
    out.close();
  }
  public static void main(String[] args) {
    new CFC();
  }

  public long[] nextLongArr(int n) {
    long[] res = new long[n];
    for(int i = 0; i < n; i++)
      res[i] = nextLong();
    return res;
  }
  public int[] nextIntArr(int n) {
    int[] res = new int[n];
    for(int i = 0; i < n; i++)
      res[i] = nextInt();
    return res;
  }
  public String nextToken() {
    while (st == null || !st.hasMoreTokens()) {
      try {
        st = new StringTokenizer(br.readLine());
      } catch (Exception e) {
        eof = true;
        return null;
      }
    }
    return st.nextToken();
  }
  public String nextString() {
    try {
      return br.readLine();
    } catch (IOException e) {
      eof = true;
      return null;
    }
  }
  public int nextInt() {
    return Integer.parseInt(nextToken());
  }
  public long nextLong() {
    return Long.parseLong(nextToken());
  }
  public double nextDouble() {
    return Double.parseDouble(nextToken());
  }
}