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

char *s = NULL;
char *t = NULL;
int l;
int foo(char *s,char *t,int i,int yn);

int main(int argc, char const *argv[]) {
  int yn = 1;
  s = malloc(sizeof(char)*200000);
  t = malloc(sizeof(char)*200000);
  scanf("%s",s);
  scanf("%s",t);
  l = strlen(s);
  yn = foo(s,t,0,1);
  if(yn == 0){
    printf("No");
  }else{
    printf("Yes");
  }
  return 0;
}


int foo(char *s,char *t,int i,int yn){
  char c,d;
  c = *(s+i);
  if(c == *(t+i)){
    for (int j = i; j < l; j++) {
      if(*(s+j) == c){
        if(*(t+j) != c){
          return 0;
        }
      }
      if(*(t+j) == c){
        if(*(s+j) != c){
          return 0;
        }
      }
    }
  }else{//別の文字の場合
    if(i+1 == l)return 0;
    d = *(t+i);
    for (int j = i; j < l; j++) {
      if(*(s+j) == c){
        *(s+j) = d;
      }else if(*(s+j) == d){
        *(s+j) = c;
      }
    }
    return foo(s,t,i,yn);
  }
  if(i+1 == l)return 1;
  return foo(s,t,i+1,yn);
}