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: Accepted)

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

typedef long long int int64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))

int func(char *s,char *t,int n){
  int i;
  char c[256];
  int used[256];
  for(i=0;i<256;i++){
    c[i]='\0';
    used[i]=0;
  }
  for(i=0;i<n;i++){
    if(c[(int)s[i]]=='\0'){
      if(used[(int)t[i]]) return 0;
      used[(int)t[i]]=1;
      c[(int)s[i]]=t[i];
    }
    if(c[(int)s[i]]!=t[i]) break;
  }
  return i==n;
}

void run(void){
  char *s=(char *)malloc(sizeof(char)*(200000+1));
  scanf("%s",s);
  char *t=(char *)malloc(sizeof(char)*(200000+1));
  scanf("%s",t);
  int n=0;
  while(s[n]!='\0') n++;
  printf("%s\n",func(s,t,n)?"Yes":"No");  
}

int main(void){
  run();
  return 0;
}