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: Time Limit Exceeded)

import java.util.Scanner;
public class Main {
	static int[][] ssame = new int[200000][2];
	static int[][] tsame = new int[200000][2];
	//static ArrayList
	static int cnt = 0,scnt = 0,tcnt = 0,sidx = 0,tidx = 0;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		String t = sc.next();
		count(s,t);
		//judge();
	}
	static void count(String s1,String s2) {
		int i,k;
		for(i = 0;i < s1.length();i++) {
			for(k = i+1;k < s1.length();k++) {
				if(s1.charAt(i) == s1.charAt(k)) {
					cnt++;
					if(s2.charAt(i) == s2.charAt(k)) {
						scnt++;
					}
				}
			}
		}
		if(cnt == scnt && scnt != 0) System.out.println("Yes");
		else {
			for(i = 0;i < s2.length();i++) {
				for(k = i+1;k < s2.length();k++) {
					if(s2.charAt(i) == s2.charAt(k)) {
						tcnt++;
					}
				}
			}
			if(cnt == 0 && tcnt == 0) System.out.println("Yes");
			else System.out.println("No");
		}
	}
	/*static void count(String s) {
		int i,k;
		for(i = 0;i < s.length();i++) {
			for(k = i+1;k < s.length();k++) {
				if(s.charAt(i) == s.charAt(k)) {
					if(cnt == 0) {
						ssame[sidx][0] = i; ssame[sidx][1] = k;
						sidx++;
					}else {
						tsame[tidx][0] = i; tsame[tidx][1] = k;
						tidx++;
					}
				}
			}
		}
	}
	static void judge() {
		if(sidx == 0 && tidx == 0 || sidx == tidx) System.out.println("Yes");
		else {
			int i,k,samecnt = 0;
			for(i = 0;i < sidx;i ++) {
				for(k = 0;k < tidx;k ++) {
					if(ssame[i][0] == tsame[k][0] && ssame[i][1] == tsame[k][1]) {
						samecnt++;
					}
				}
			}
			if(samecnt == sidx && samecnt != 0) System.out.println("Yes");
			else System.out.println("No");
		}
	}*/
}