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 <iostream>
#include <cmath>
#include <string>
#include <algorithm>
#include <set>
#include <vector>
#include <map>
#include <list>
#include <stack>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#define pye cout << "yes" << endl
#define pno cout << "no" << endl
#define pYe cout << "Yes" << endl
#define pNo cout << "No" << endl
#define pYE cout << "YES" << endl
#define pNO cout << "NO" << endl
#define prin(num) cout << num << endl
#define max(a, b) ((a)>(b) ? (a):(b))
#define min(a, b) ((a)<(b) ? (a):(b))
#define INF 1000000000
#define MOD 1000000007LL
#define MAX 10005
#define MAX_E 30005
#define NIL -1
#define WHITE 0
#define BLACK 1
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef map<ll, ll> Map;

int main(void) {
    string s, t;
    cin >> s >> t;
    ll i;
    ll pair[26];
    
    for(i=0; i<26; i++)
        pair[i]=-1;
    
    for(i=0; i<s.size(); i++) {
        if(s[i]==t[i] && pair[s[i]-'a']==-1) {
            pNo;
            return 0;
        }
        else if(s[i]==t[i] && pair[s[i]-'a']!=-1) {
            continue;
        }
        else if(pair[s[i]-'a']!=-1) {
            s[i]=(char)(pair[s[i]-'a']+'a');
            if(s[i]!=t[i]) {
                pair[s[i]-'a']=t[i]-'a';
                pair[t[i]-'a']=s[i]-'a';
            }
        }
        else if(pair[s[i]-'a']==-1) {
            pair[s[i]-'a']=t[i]-'a';
            pair[t[i]-'a']=s[i]-'a';
        }
    }
    
    pYe;
}