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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define lper(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[4]={1,0,-1,0};
const ll dx[4]={0,-1,0,1};
int main(){
    string S,T;cin>>S>>T;
    vector<int> R(26,-1),r(26,-1);
    bool check=0;
    rep(i,S.size()){
        if(R[S[i]-'a']==-1)R[S[i]-'a']=T[i]-'a';
        else{
            if(R[S[i]-'a']!=T[i]-'a'){
                check=1;
                break;
            }
        }
    }
    rep(i,T.size()){
        if(r[T[i]-'a']==-1)r[T[i]-'a']=S[i]-'a';
        else{
            if(r[T[i]-'a']!=S[i]-'a'){
                check=1;
                break;
            }
        }
    }
    if(check)cout<<"No"<<endl;
    else cout<<"Yes"<<endl;
    return 0;
}