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>

#define NMAX 200000

main()
{
  char S[NMAX],T[NMAX];
  int alp[26],i,num,k;
  for(i=0;i<26;i++)
    alp[i] = -1;
  scanf("%s",S);
  scanf("%s",T);
  for(num=0;S[num]!='\0';num++);
  for(i=0; i<num; i++)
  {
    if(alp[S[i]-97]==-1)
      alp[S[i]-97]=T[i]-97;
    else
    {
      if(alp[S[i]-97]!=T[i]-97)
      {
        printf("No\n");
        exit(0);
      }
    }
  }
  for(i=0; i<26; i++)
  {
    num = alp[i];
    if(num!=-1&&i!=num)
    {
      if(alp[num]==num)
        {
          printf("No\n");
          exit(0);
        }
        
    }
    if(i<25&&num!=-1)
    {
      for(k=i+1; k<26; k++)
      {
        if(num == alp[k])
        {
          printf("No\n");
          exit(0);
        }
      }
    }
  }
printf("Yes\n");
}