You are given strings S and T consisting of lowercase English letters.

You can perform the following operation on S any number of times:

Operation: Choose two distinct lowercase English letters c_1 and c_2, then replace every occurrence of c_1 with c_2, and every occurrence of c_2 with c_1.

Determine if S and T can be made equal by performing the operation zero or more times.

Constraints

* 1 \leq |S| \leq 2 \times 10^5
* |S| = |T|
* S and T consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:


S
T


Output

If S and T can be made equal, print `Yes`; otherwise, print `No`.

Examples

Input

azzel
apple


Output

Yes


Input

chokudai
redcoder


Output

No


Input

abcdefghijklmnopqrstuvwxyz
ibyhqfrekavclxjstdwgpzmonu


Output

Yes
/*
 * Click `Run` to execute the snippet below!
 */

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

/*
 * To execute Java, please define "static void main" on a class
 * named Solution.
 *
 * If you need more classes, simply define them inline.
 */

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    String s = sc.next(), t = sc.next();
    int N = s.length();
    TreeMap<Character, Character> StoT = new TreeMap<>();
    TreeMap<Character, Character> TtoS = new TreeMap<>();
    for(int i = 0; i<N; i++){
      //First Observation for every character in S they can only map to one unique character of T
      if(StoT.containsKey(s.charAt(i)) && StoT.get(s.charAt(i)) != t.charAt(i)){
        System.out.println("No");
        System.exit(0);
      }
      //Second Observation for every character in T they can only map to one unique character of S
      if(TtoS.containsKey(t.charAt(i)) && TtoS.get(t.charAt(i)) != s.charAt(i)){
        System.out.println("No");
        System.exit(0);
      }   
      StoT.put(s.charAt(i), t.charAt(i));
      TtoS.put(t.charAt(i), s.charAt(i));
    }
    System.out.println("Yes");
  }
}
