Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

azzel
apple

Sample Output 1

Yes

azzel can be changed to apple, as follows:

Choose e as c_1 and l as c_2. azzel becomes azzle.

Choose z as c_1 and p as c_2. azzle becomes apple.

Sample Input 2

chokudai
redcoder

Sample Output 2

No

No sequences of operation can change chokudai to redcoder.

Sample Input 3

abcdefghijklmnopqrstuvwxyz
ibyhqfrekavclxjstdwgpzmonu

Sample Output 3

Yes

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        String S  = sc.nextLine();
        String T = sc.nextLine();

        HashMap<Character, Character> map = new HashMap<>();
        Set<Character> li = new HashSet<>();

        for (int i = 0; i < S.length(); i++) {
            if (S.charAt(i) == T.charAt(i)) {
                li.add(S.charAt(i));
                continue;
            }

            if (map.containsKey(S.charAt(i)) && map.get(S.charAt(i)) != T.charAt(i)) {
                System.out.println("No");
                return;
            }

            map.put(S.charAt(i), T.charAt(i));
        }

        for (int i = 0; i < S.length(); i++) {
            if (S.charAt(i) != T.charAt(i) && li.contains(T.charAt(i))) {
                System.out.println("No");
                return;
            }
        }

        System.out.println("Yes");
    }
}