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>

using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, k, n) for (int i = (int)(k); i < (int)(n); i++)
#define rrep(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define rREP(i, k, n) for (int i = (int)(n)-1; i >= k; i--)

#define fi first
#define se second

#define vi vector<int>
#define pb push_back
#define mp make_pair

#define pcnt __builtin_popcount

typedef long long ll;

const int inf = 100100100;
const int MOD = (int) 1e9 + 7;
const double EPS = 1e-9;

template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; }return 0; }

template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b;return 1; }return 0; }

int main() {
    string s,t;
    cin >> s;
    cin >> t;
    int arr[27]={};
    rep(i,27) arr[i] = -1;
    rep(i,s.size()){
        if(arr[s[i]-'a']<0){
            arr[s[i]-'a'] = t[i]-'a';
        }else{
            if(arr[s[i]-'a']!=arr[t[i]-'a']){
                cout << "No" << endl;
                return 0;
            }
        }
    }
    cout << "Yes" << endl;
    return 0;
}