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 <stdlib.h>
#include <string.h>
#include <math.h>
int main(void){
  int i;
  char s[200000],t[200000];
  char R1[30],R2[30];
  for(i=0;i<30;i++){
    R1[i]=-1;R2[i]=-1;
  }
  scanf("%s",s);
  scanf("%s",t);
  for(i=0;i<strlen(s);i++){
    if(R1[s[i]-'a']>0){
      if(t[i]!=R1[s[i]-'a']){
	printf("No\n");
	return 0;
      }
    }
    if(R2[t[i]-'a']>0){
      if(R2[t[i]-'a']!=s[i]){
	printf("No\n");
	return 0;
      }
    }
    if(R1[s[i]-'a']<0)
      R1[s[i]-'a']=t[i];
    if(R2[t[i]-'a']<0)
      R2[t[i]-'a']=s[i];
  } 
  printf("Yes\n");
  return 0;
}