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: Wrong Answer)

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

int main(){
  char *s = NULL;
  char *t = NULL;
  ssize_t length;

  if((length = getline(&s, &(size_t){4096}, stdin))== EOF){
    return 1;
  }
  if(s[length - 1] == '\n'){
    s[length--] = '\0';
  }
  if((length = getline(&t, &(size_t){4096}, stdin))== EOF){
    return 1;
  }
  if(t[length - 1] == '\n'){
    t[length--] = '\0';
  }

  //  printf("%s", s);
  //  printf("%s", t);

  if(strcmp(s, t) == 0){
    printf("Yes\n");
    return 0;
  }

  int i, j;
  char tmp;
  for(i = 0; i < length; i++){
    for(j = i + 1; j < length; j++){
      if(s[i] > s[j]){
	tmp = s[i];
	s[i] = s[j];
	s[j] = tmp;
      }
    }
  }
  for(i = 0; i < length; i++){
    for(j = i + 1; j < length; j++){
      if(t[i] > t[j]){
	tmp = t[i];
	t[i] = t[j];
	t[j] = tmp;
      }
    }
  }

  for(i = 0; i < length; i++){
    for(j = i + 1; j < length; j++){
      if(s[i] == s[j]){
	if(t[i] == t[j]){
	  s[i] = t[i];
	  s[j] = t[j];
	}
      }
    }
  }

  // printf("%s", s);
  //printf("%s", t);

  if(strcmp(s, t) == 0){
    printf("Yes\n");
  }
  else{
    printf("No\n");
  }

  free(s);
  free(t);

  return 0;
}