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>
//小さい順にソート
char c[2][200002];
void merge(int i,int j,char s[2][200002]){
  if(i!=j){
    merge(i,(i+j)/2,s);
    merge((i+j)/2+1,j,s);
    int a=i,b=(i+j)/2+1;
    int d;
    for(d=i;d<=j;d++){
      if((i+j)/2<a){
        c[0][d]=s[0][b];
        c[1][d]=s[1][b];
        b++;
      }
      else if(j<b){
        c[0][d]=s[0][a];
        c[1][d]=s[1][a];
        a++;
      }
      else if(s[0][a]<s[0][b]){
        c[0][d]=s[0][a];
        c[1][d]=s[1][a];
        a++;
      }
      else{
        c[0][d]=s[0][b];
        c[1][d]=s[1][b];
        b++;
      }
    }
    for(d=i;d<=j;d++){
      s[0][d]=c[0][d];
      s[1][d]=c[1][d];
    }
  }
}

int main(){
  char s[2][200002];
  scanf("%s",s[0]);
  scanf("%s",s[1]);
  int len,i,change[27];
  for(i=0;i<27;i++)
    change[i]=0;
  len=strlen(s[0]);
  merge(0,len-1,s);
  for(i=0;i<len;i++)
    if(s[0][i]==s[1][i])
      change[s[0][i]-'a']=1;
  for(i=0;i<len-1;i++)
    if((s[0][i]==s[0][i+1] && s[1][i]!=s[1][i+1]) || (s[0][i]!=s[1][i] && (change[s[1][i]-'a']==1 || change[s[0][i]-'a']==1))){
      puts("No");
      return 0;
    }
  puts("Yes");
  return 0;
}