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: Wrong Answer)

#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <string>
#include <cmath>

using namespace std;

int ci(char s){

	if(s=='a') return 0;
	if(s=='b') return 1;
	if(s=='c') return 2;
	if(s=='d') return 3;
	if(s=='e') return 4;
	if(s=='f') return 5;
	if(s=='g') return 6;
	if(s=='h') return 7;
	if(s=='i') return 8;
	if(s=='j') return 9;
	if(s=='k') return 10;
	if(s=='l') return 11;
	if(s=='m') return 12;
	if(s=='n') return 13;
	if(s=='o') return 14;
	if(s=='p') return 15;
	if(s=='q') return 16;
	if(s=='r') return 17;
	if(s=='s') return 18;
	if(s=='t') return 19;
	if(s=='u') return 20;
	if(s=='v') return 21;
	if(s=='w') return 22;
	if(s=='x') return 23;
	if(s=='y') return 24;
	if(s=='z') return 25;

	cout<< "ci error" <<endl;
}

int main(){

	int tmp;
	string s, t;
	vector<int> tc(26, 0);

	cin>> s >> t;

	for(int i=0;i<t.size();i++){
		tmp=ci(t[i]);
		if(tc[tmp]==0){
			tc[tmp]=26+tmp-ci(s[i]);
		}else{
			if(tc[tmp]!=26+tmp-ci(s[i])){
				cout<< "No" << endl;
				return 0;
			}
		}
	}

	cout<< "Yes" << endl;


}