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 <iostream>
#include <algorithm>
#include <string>
using namespace std;
int main(){
	string a;
	string b;
	int i;
	cin >> a >> b;
	int tong[26];
	int used[26];
	for ( i = 0 ; i < 26 ; i++ ){
		tong[i] = -1;
		used[i] = 0;
	}
		
	for ( i = 0 ; i < a.length(); i ++){
		if ( used[a[i] - 'a'] && tong[a[i] - 'a'] != b[i] - 'a'){
			cout << "No";
			return 0;
		}
		tong[a[i] - 'a'] = b[i] - 'a';
		used[a[i] - 'a'] = 1;
	}
	for ( i = 0 ; i < 26 ; i++ ){
		tong[i] = -1;
		used[i] = 0;
	}
		
	for ( i = 0 ; i < a.length(); i ++){
		if ( used[b[i] - 'a'] && tong[b[i] - 'a'] != a[i] - 'a'){
			cout << "No";
			return 0;
		}
		tong[b[i] - 'a'] = a[i] - 'a';
		used[b[i] - 'a'] = 1;
	}
	cout << "Yes";
}