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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <limits.h>

int main(void){
    char s[200020],t[200020];
    scanf("%s%s",s,t);
    int x[27],y[27];
    for (int i=0; i<27; i++){
        x[i]=-1;
        y[i]=-1;
    }
    for (int i=0; i<strlen(s); i++){
        if (x[s[i]-'a']==-1){
            x[s[i]-'a']=t[i]-'a';
        } else {
            if (x[s[i]-'a']!=t[i]-'a'){
                printf("No\n");
                return 0;
            }
        }
    }
    for (int i=0; i<strlen(t); i++){
        if (y[t[i]-'a']==-1){
            y[t[i]-'a']=s[i]-'a';
        } else {
            if (y[t[i]-'a']!=s[i]-'a'){
                printf("No\n");
                return 0;
            }
        }
    }
    printf("Yes\n");
    return 0;
}