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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long int
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
static const ll MAX = 1000000000000000;
static const int NIL = -1;
using namespace std;
const long double EPS = 0.0000000001;
const long double PI = (acos(-1));
const int MOD = 1000000007;

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")




int main() {


    string s,t; cin >> s >> t;
    ll n = s.length();
    map<char,ll> mp;
    rep(i,26){
        mp[(char)('a'+i)] = 0;
    }
    rep(i,n){
        if(t[i] != s[i]){
            if(mp[t[i]] == 0){
                mp[t[i]] = (int)(s[i]-t[i]);
            }else{
                if(mp[t[i]] != (int)(s[i]-t[i])){
                    cout << "No";
                    return 0;
                }
            }
        }
    }

    rep(i,26){
        mp[(char)('a'+i)] = 0;
    }
    rep(i,n){
        if(t[i] != s[i]){
            if(mp[s[i]] == 0){
                mp[s[i]] = (int)(s[i]-t[i]);
            }else{
                if(mp[s[i]] != (int)(s[i]-t[i])){
                    cout << "No";
                    return 0;
                }
            }
        }
    }

    cout << "Yes";
    return 0;
}