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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair< ll, ll > Pi;
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep2(i,n) for(int i=1;i<=(n);i++)
#define rep3(i,i0,n) for(int i=i0;i<(n);i++)
#define pb push_back
#define mod 1000000007
const ll INF = 1LL << 60;
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; }
ll gcd(ll a, ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) {return a/gcd(a,b)*b;}
#define all(x) x.begin(), x.end()
#define mp make_pair
bool compare(Pi a, Pi b) {
    if(a.first != b.first){
        return a.first < b.first;
        
    }else{
        return a.second < b.second;
    }
}


bool In_map(ll y,ll x,ll h,ll w){
    if(y<0 || x<0 || y>=h || x>=w){
        return 0;
    }else{
        return 1;
    }
}
const vector<ll> dx{1,0,-1,0};
const vector<ll> dy{0,1,0,-1};


int main() {
    string S,T;
    cin >>S>>T;
    map<char,ll>m1;
    map<char,ll>m2;
    rep(i,S.size()){
        if(m1[S[i]]!=m2[T[i]]){
            cout<<"No"<<endl;
            return 0;
        }
        m1[S[i]] =i+1;
        m2[T[i]] =i+1;
    }
    cout << "Yes" << endl;


    return 0;
    

}