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.HashMap;
import java.util.Map;
import java.util.Scanner;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		String t = sc.next();
		Map<String, Integer> smap = new HashMap<>();
		Map<String, Integer> tmap = new HashMap<>();
		
		for (int i = 0; i < s.length(); i++) {
			if (smap.containsKey(String.valueOf(s.charAt(i)))) {
				smap.put(String.valueOf(s.charAt(i)), smap.get(String.valueOf(s.charAt(i))+1));
			} else {
				smap.put(String.valueOf(s.charAt(i)), 1);
			}
		}
		for (int i = 0; i < t.length(); i++) {
			if (tmap.containsKey(String.valueOf(t.charAt(i)))) {
				tmap.put(String.valueOf(t.charAt(i)), tmap.get(String.valueOf(t.charAt(i))+1));
			} else {
				tmap.put(String.valueOf(t.charAt(i)), 1);
			}
		}
		
		if (smap.size() != tmap.size()) {
			System.out.println("No");
		} else {
			System.out.println("Yes");
		}
	}	
}