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.Arrays;
import java.util.Scanner;

public class Main {

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

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

		int len = S.length();
		char[] cs = S.toCharArray();
		char[] ct = T.toCharArray();

		char[] cs2 = S.toCharArray();
		char[] ct2 = T.toCharArray();

		Arrays.sort(cs2);
		Arrays.sort(ct2);

		// 同じ文字を縮めた時の長さを数える
		int s1=1;
		int t1=1;
		boolean b=true;
		for (int i=1;i<len;i++){
			if(cs[i]!=cs[i-1]) {s1++;}
			if(ct[i]!=ct[i-1]) {t1++;}
			if (s1!=t1) {
				b = false;
			}
		}

		// 文字の種別数を数える
		int s2=1;
		int t2=1;
		for (int i=1;i<len;i++){
			if(cs2[i]!=cs2[i-1]) {s2++;}
			if(ct2[i]!=ct2[i-1]) {t2++;}
		}

		if (b && (s2==t2)) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}

	}
}