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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

#define STRING_MAX (200000+1)
#define MAP_SIZE 26
#define A_VALUE 97

int main(void){
    char s[STRING_MAX], t[STRING_MAX];
    int smap[MAP_SIZE], tmap[MAP_SIZE];
    for(int i = 0; i < MAP_SIZE; i++){
        smap[i] = -1;
        tmap[i] = -1;
    }
    scanf("%s\n%s", s, t);
    int length = strlen(s);

    for(int i = 0; i < length; i++){
        if(smap[t[i] - A_VALUE] < 0){
            smap[t[i] - A_VALUE] = t[i];
        }
        else{
            if(smap[t[i] - A_VALUE] != t[i]){
                printf("No\n");
                return 0;
            }
            if(tmap[s[i] - A_VALUE] != s[i]){
                printf("No\n");
                return 0;
            }
        }
        if(tmap[s[i] - A_VALUE] < 0){
            tmap[s[i] - A_VALUE] = s[i];
        }
        else{
            if(tmap[s[i] - A_VALUE] != s[i]){
                printf("No\n");
                return 0;
            }
            if(smap[t[i] - A_VALUE] != t[i]){
                printf("No\n");
                return 0;
            }
        }
    }
    printf("Yes\n");
    return 0;
}