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: Runtime Error)

#include <stdio.h>
int main(void){
   char a[2*100000];
   char b[2*100000];
   int j,i;
   
   scanf("%s",a);
   scanf("%s",b);
   
 for(i=0;a[i]!='\0';i++){
     if(a[i]==b[i]){
        for(j=i+1;b[j]!='\0';j++){
                 if(b[i]==b[j]){
                     if(b[j]!=a[j]) {
                         printf("No");
                         return 0;
                     }
                     
                 }
         }
     }
 }
 for(i=0;b[i]!='\0';i++){
     if(b[i]==a[i]){
        for(j=i+1;a[j]!='\0';j++){
                 if(a[i]==a[j]){
                     if(a[j]!=b[j]) {
                         printf("No");
                         return 0;
                     }
                     
                 }
         }
     }
 }
   
  printf("Yes");
  return 0;
}