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: Accepted)

import java.util.*;
import java.lang.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        String t = sc.next();
        int a[] = new int[26];
        int b[] = new int[26];
        boolean bre = false;
        for (int i = 0; i < 26; i++) {
            a[i] = -1;
            b[i] = -1;
        }
        for (int i = 0; i < s.length(); i++){
            if (a[s.charAt(i) - 'a'] >= 0) {
                if (a[s.charAt(i) - 'a'] != t.charAt(i) - 'a') {
                    bre = true;
                    break;
                }
            }
            if (b[t.charAt(i) - 'a'] >= 0) {
                if (b[t.charAt(i) - 'a'] != s.charAt(i) - 'a') {
                    bre = true;
                    break;
                }
            }
            if (a[s.charAt(i) - 'a'] < 0) {
                a[s.charAt(i) - 'a'] = t.charAt(i) - 'a';
            }
            if (b[t.charAt(i) - 'a'] < 0) {
                b[t.charAt(i) - 'a'] = s.charAt(i) - 'a';
            }
        }
        if (!bre) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}