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 <string.h>

int main(void){
    char S[200001];
    char T[200001];
    memset(&S,'\0',sizeof(S));
    memset(&T,'\0',sizeof(T));
    scanf("%s",S);
    scanf("%s",T);
    
    int size = strlen(S);

    char alpha[26];
    memset(&alpha,'\0',sizeof(alpha));
    for(int i = 0;i<size;i++){
        // printf("alpha %c\n",alpha[T[i] - 'a']);
        // printf("alalpha %c\n",alpha[alpha[T[i] - 'a'] - 'a']);
        if(alpha[T[i] - 'a'] == '\0'){
            alpha[T[i] - 'a'] = S[i];
        }else if(alpha[T[i] - 'a'] != S[i]){
            printf("No\n");
            return 0;
        }else{
            continue;
        }
        // if(alpha[T[i] - 'a'] == '\0'){
        //     alpha[T[i] - 'a'] = S[i];
        // }
        // if(alpha[S[i] - 'a'] == '\0'){
        //     alpha[S[i] - 'a'] = T[i];
        // }
        // printf("al t%c\n",alpha[T[i] - 'a']);
        // printf("al s%c\n",alpha[S[i] - 'a']);
        // printf("s %c\n",S[i]);
        // printf("t %c\n",T[i]);
        // if(alpha[T[i] - 'a'] != S[i]){
        //     printf("No\n");
        //     return 0;
        // }
        // if(alpha[S[i] - 'a'] != T[i]){
        //     printf("No\n");
        //     return 0;
        // }
        // if(alpha[S[i] - 'a'] == '\0'){
        //     alpha[S[i] - 'a'] = T[i];
        // }else if(alpha[S[i] - 'a'] != T[i]){
        //     printf("No\n");
        //     return 0;
        // }else{
        //     continue;
        // }
    }
    // printf("S %s\n",S);
    // printf("T %s\n",T);
    printf("Yes\n");
    return 0;
}