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: Wrong Answer)

#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i++)

int main(void){
	char s[200000],t[200000];
	scanf("%s",s);
	scanf("%s",t);
	int i,j,flag=0;
	int len=strlen(s);
	char R[26],Rd[26];
	rep(i,26){
		R[i]=-1;
		Rd[i]=-1;
	}
	for(i=0;i<len;i++){
		flag=0;
		if(R[s[i]-97]>=0){
			if(R[s[i]-97]!=t[i]){
				printf("No");
				return 0;
			}
		} 
		if(Rd[t[i]-97]>=0){
			if(Rd[t[i]-97]!=s[i]){
				printf("No");
				return 0;
			}
		}
		if(R[s[i]-97]<0) R[s[i]-97]=t[i];
		if(Rd[t[i]-97]<0) Rd[t[i]-97]=s[i];
	}


	printf("Yes");
	return 0;
}