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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>

int main(void){
	char    S[200001];
	char    T[200001];
	int i,j;
	int R0[26];
	int R1[26];
	for(i=0;i<26;i++){
		R0[i]=-1;
		R1[i]=-1;
	}
	scanf("%s",S);
	scanf("%s",T);
	for(i=0;i<strlen(S);i++){
		if(R0[S[i]-'a']<0){
			R0[S[i]-'a'] = T[i]-'a';
		}
		if(R1[T[i]-'a']<0){
			R1[T[i]-'a'] = S[i]-'a';
		}
	}
	for(i=0;i<strlen(S);i++){
		if(R0[S[i]-'a']!=T[i]-'a'){
			printf("No");
			return 0;
		}
		if(R1[T[i]-'a']!=S[i]-'a'){
			printf("No");
			return 0;
		}
	}
	printf("Yes");
	return 0;
}