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 Java7 (OpenJDK 1.7.0) (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);
		Main tr = new Main();
		tr.test(sc);
		sc.close();
	}

	void test(Scanner sc){
		char[] S = sc.nextLine().toCharArray();
		char[] T = sc.nextLine().toCharArray();
		if(S.length==1){
			System.out.println("Yes");return;
		}

		Arrays.sort(S);
		Arrays.sort(T);
		int[] fS = new int[26];
		char p = S[0];
		int c=1;
		int sidx = 0;
		for (int i = 1; i < S.length; i++) {
			if(p==S[i]){
				c++;
			}else{
				fS[sidx++] = c;
				c=1;
			}
			p=S[i];
		}


		int[] fT = new int[26];
		p = T[0];
		c=1;
		int tidx = 0;
		for (int i = 1; i < T.length; i++) {
			if(p==T[i]){
				c++;
			}else{
				fT[tidx++] = c;
				c=1;
			}
			p=T[i];
		}

		if(tidx != sidx){
			System.out.println("No");
			return;		
		}
		Arrays.sort(fS, 0, sidx);
		Arrays.sort(fT, 0, tidx);
		for (int i = 0; i < sidx; i++) {
			if(fS[i]!= fT[i]){
				System.out.println("No");
				return;
			}
		}


		System.out.println("Yes");


	}
}