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: Time Limit Exceeded)

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

int main(){
  int i,j,l;
  int k=0;
  int r;
  char s[256000];
  char t[256000];
  char p;
  char q;

  scanf("%s",s);
  scanf("%s",t);

  r = strlen(s);

  for(i=r-1; i>=0; i--){
    if(s[i]!=t[i]){
      l=i;
      p=s[i];
      q=t[i];
      for(j=r-1; j>=0; j--){
        if(s[j]==q){
          s[j]=p;
        }else if(s[j]==p){
          s[j]=q;
        }
      }
    }
  }

  for(i=0; i<=r-1; i++){
    if(s[i]!=t[i]){
      k++;
    }
  }
  //printf("%d\n",k);
  if(k==0){
    printf("Yes\n");
  }else{
    printf("No\n");
  }

  return 0;
}