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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.util.Objects;
import java.util.StringTokenizer;

public class CodeforceTemplate {


  private static FastScanner scanner = new FastScanner();
  private static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));

  public static void main(String[] args) {
    int tests = scanner.nextInt();
    for (int i = 0; i < tests; i++) {
      solve();
    }
  }

  private static void solve() {
    int N = rInt();
    int K = rInt();
    String start = rStr();
    String target = rStr();
    char[] a = start.toCharArray();
    char[] b = target.toCharArray();
    Arrays.sort(a);
    Arrays.sort(b);
    for (int i = 0; i < N; i++) {
      if (a[i] > b[i]) {
        wl("No");
        return;
      }
    }
    int[] charsA = new int[26];
    int[] charsB = new int[26];
    for (int i = 0; i < N; i++) {
      charsA[a[i] - 'a']++;
      charsB[b[i] - 'a']++;
    }
    for (int i = 0; i < 26; i++) {
      if (Math.abs(charsA[i] - charsB[i]) % K != 0) {
        wl("No");
        return;
      }
    }
    wl("Yes");
  }

  private static final int[][] DIRS_2D = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};

  private final static class FastScanner {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String next() {
      while (!st.hasMoreTokens()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }

    long nextLong() {
      return Long.parseLong(next());
    }

    int[] readIntArray(int size) {
      int[] arr = new int[size];
      for (int i = 0; i < size; i++) {
        arr[i] = scanner.nextInt();
      }
      return arr;
    }

    long[] readLongArray(int size) {
      long[] arr = new long[size];
      for (int i = 0; i < size; i++) {
        arr[i] = scanner.nextLong();
      }
      return arr;
    }
  }

  private static void wl(String ans) {
    try {
      out.write(ans);
      out.newLine();
      out.flush();
    } catch (IOException e) {
      e.printStackTrace();
    }
  }

  private static void wl(long ans) {
    wl(String.valueOf(ans));
  }


  private static void wl(int ans) {
    wl(String.valueOf(ans));
  }

  private static int rInt() {
    return scanner.nextInt();
  }

  private static long rLong() {
    return scanner.nextLong();
  }

  private static String rStr() {
    return scanner.next();
  }

  private static int[] rIntArr(int size) {
    return scanner.readIntArray(size);
  }

  private static long[] rLongArr(int size) {
    return scanner.readLongArray(size);
  }

  private static class Pair<U, V> implements Comparable<Pair<U, V>> {

    final U first;
    final V second;

    static <U, V> Pair<U, V> makePair(U first, V second) {
      return new Pair<U, V>(first, second);
    }

    private Pair(U first, V second) {
      this.first = first;
      this.second = second;
    }

    @Override
    public boolean equals(Object o) {
      if (this == o) {
        return true;
      }
      if (o == null || getClass() != o.getClass()) {
        return false;
      }

      Pair pair = (Pair) o;

      return Objects.equals(first, pair.first) &&
          Objects.equals(second, pair.second);
    }

    @Override
    public int hashCode() {
      int result = first != null ? first.hashCode() : 0;
      result = 31 * result + (second != null ? second.hashCode() : 0);
      return result;
    }

    public Pair<V, U> swap() {
      return makePair(second, first);
    }

    @Override
    public String toString() {
      return "(" + first + "," + second + ")";
    }

    @SuppressWarnings({"unchecked"})
    public int compareTo(Pair<U, V> o) {
      int value = ((Comparable<U>) first).compareTo(o.first);
      if (value != 0) {
        return value;
      }
      return ((Comparable<V>) second).compareTo(o.second);
    }
  }
}