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

char str1[200001],str2[200001];
int main(){
int len,i,j;
int alphabets[26],checkcount = 0,error = 0;
for(i = 0;i < 26;i++){
alphabets[i] = -1;
}
scanf("%s",str1);
scanf("%s",str2);
len = strlen(str1);

for(i = 0;i < len;i++){
if(alphabets[str1[i] - 'a'] == -1){
alphabets[str1[i] - 'a']= str2[i] - 'a';
}
else if(alphabets[str1[i] - 'a'] != str2[i] -'a'){error = 1;break;}
}

for(i = 0;i < 26;i++){
checkcount = 0;
for(j = 0;j < 26;j++){
if(alphabets[j] == i){checkcount++;}
}
if((alphabets[i] != -1)!=checkcount){error = 1;}
}
if(error == 0){printf("Yes\n");}else{printf("No\n");}
return 0;
}