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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
using namespace std;
#define L 200005

char s[L],t[L];
int len,last[30];
int lasts[L],lastt[L];
bool flag=true;

int main() {
	scanf("%s %s",s,t);
	len=strlen(s);
	for(int i=0;i<len;i++) {
		int j=s[i]-'a';
		lasts[i]=last[j];
		last[j]=i;
	}
	memset(last,0,sizeof(last));
	for(int i=0;i<len;i++) {
		int j=t[i]-'a';
		lastt[i]=last[j];
		last[j]=i;
	}
	for(int i=0;i<len;i++)
		if(lasts[i]!=lastt[i]) {
			printf("No\n");
			return 0;
		}
	printf("Yes\n");
	return 0;
}