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.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.concurrent.ConcurrentHashMap;

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

		Map<Character, Integer> sCount = new ConcurrentHashMap<>();
		Map<Character, Integer> tCount = new ConcurrentHashMap<>();
		for (int i = 0; i < S.length(); ++i) {
			Character c = S.charAt(i);
			if (!sCount.containsKey(c)) {
				sCount.put(c, 1);
			} else {
				sCount.put(c, sCount.get(c) + 1);
			}
		}
		for (int i = 0; i < T.length(); ++i) {
			Character c = T.charAt(i);
			if (!tCount.containsKey(c)) {
				tCount.put(c, 1);
			} else {
				tCount.put(c, tCount.get(c) + 1);
			}
		}
		List<Integer> sList = new ArrayList<>(sCount.size());
		List<Integer> tList = new ArrayList<>(tCount.size());
		sList.addAll(sCount.values());
		tList.addAll(tCount.values());
		Collections.sort(sList);
		Collections.sort(tList);
		boolean result = true;
		if (sList.size() != tList.size()) {
			result = false;
		} else {
			for (int i = 0; i < sList.size(); ++i) {
				if (sList.get(i) != tList.get(i)) {
					result = false;
					break;
				}
			}
		}
		if (result) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}
		in.close();
	}
}