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 (Clang 3.8.0) (judge verdict: Accepted)

#if defined(LOCAL)
const double _max_double_error = 1e-9;
#include "testutils.h"
#define L(x...) (debug(x, #x))
#define I(x, ...) (x)
#define C(x...) CHECK(x)
#else
#define L(x, ...) (x)
#define I(x, ...) (x)
#define C(x, ...) ;
#endif
#include <algorithm>
#include <vector>
#include <string>
#include <iomanip>
#include <iostream>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <math.h>
#include <limits>
#include <numeric>
#include <queue>
#include <memory>

using namespace std;
using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using ii = pair<int,int>; using lu = unsigned long long; using l = long long;
using vs = vector<string>; using vii = vector<ii>;
using vl = vector<l>; using vvl = vector<vl>; using vvvl = vector<vvl>;
using ll = pair<l,l>; using vll = vector<ll>; using vvll = vector<vll>;
using vb = vector<bool>; using vvb = vector<vb>;
using vd = vector<double>; using vvd = vector<vd>;
using mll = unordered_map<l, l>; using sl = unordered_set<l>;
const l INF = numeric_limits<l>::max();
const double EPS = 1e-10; const double PI = M_PI;
const l e0=1, e3=1000, e5=100000, e6=10*e5, e7=10*e6, e8=10*e7, e9=10*e8;
const char lf = '\n';
#define all(x) begin(x), end(x)
#define F(a,b,c) for (l a = l(b); a < l(c); a++)
#define B(a,b,c) for (l a = l(c) - 1; a >= l(b); a--)
#define VVL(x, a, b, i) vvl x(a, vl(b, l(i)));
#define VVVL(x, a, b, c, i) vvvl x(a, vvl(b, vl(c, l(i))));

void solve(istream& in, ostream& out);
int main(int argc, char **argv) {
  ios_base::sync_with_stdio(false); cin.tie(0);
  cout << fixed << setprecision(15);
#if defined(LOCAL)
  tst::test_init(argc, argv);
  // _generate_random_test = generate_random;
  // _solve_brute = solve_brute;
  // _player_b = player_b;
  // _custom_solution_checker = solution_checker;
  tst::maybe_run_tests(cin, cout);
#else
  solve(cin, cout);
#endif
}
const l MOD = e9 + 7; // end of template

void solve(istream& in, ostream& out) {
  string s, t; in >> s >> t;
  vl outgoing(26, -1), incoming(26, -1);
  bool ok = true;
  F(i, 0, s.size()) {
    l a = s[i] - 'a', b = t[i] - 'a';
    ok = ok and (outgoing[a] == -1 or outgoing[a] == b) and
      (incoming[b] == -1 or incoming[b] == a);
    outgoing[a] = b;
    incoming[b] = a;
  }
  out << (ok ? "Yes" : "No") << lf;
}