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 DEBUG(x) cout<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cout<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl


typedef long long ll;
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define rrep1(i,n) for(int i=n;i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(v) v.begin(),v.end()
#define UNIQUE(v) v.erase(unique(all(v)), v.end())
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 (a>b) { a = b; return 1; } return 0; }
const ll inf = 1000000001;
const ll INF = 1e18;
const ll MOD = 1000000007;
const double pi = 3.14159265358979323846;
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { 1,0, -1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
//#define mp make_pair

int main() {
	string s, t;
	cin >> s >> t;
	vector<char> to(26, '?');
	rep(i, s.size()) {
		if (to[s[i] - 'a'] == '?') {
			to[s[i] - 'a'] = t[i];
		}
		else {
			if (to[s[i] - 'a'] != t[i]) {
				cout << "No" << endl;
				return 0;
			}
		}
	}
	vi used(26, 0);
	rep(i, 26) {
		if (to[i] == '?') {
			continue;
		}
		if (used[to[i] - 'a']) {
			cout << "No" << endl;
			return 0;
		}
		else {
			used[to[i] - 'a'] = 1;
		}
	}
	cout << "Yes" << endl;
	return 0;
}