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 IOS ios::sync_with_stdio(false); cin.tie(0);
#define FOR(i,s,n) for(int i = (s); i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) for(int i = (n); i >= 0; i--)
#define ALL(n) (n).begin(), (n).end()
#define RALL(n) (n).rbegin(), (n).rend()
#define ATYN(n) cout << ( (n) ? "Yes":"No") << '\n';
#define CFYN(n) cout << ( (n) ? "YES":"NO") << '\n';
#define OUT(n) cout << (n) << '\n';
using ll = long long;
using ull = unsigned long long;
using pii = pair<int,int>;
using pll = pair<ll,ll>;

int main(void)
{
    IOS
    string s; cin >> s;
    string t; cin >> t;
    const int N = 26;
    vector<int> sp(N,-1);
    vector<int> tp(N,-1);
    auto apos = [](char c) {return c - 'a';};

    int si = 0, ti = 0;
    bool ans = true;
    REP(i,s.size()) {
        int sc = apos(s[i]);
        int tc = apos(t[i]);
        if (sp[sc] == -1) sp[sc] = si++;
        if (tp[tc] == -1) tp[tc] = ti++;
        if (sp[sc] !=tp[tc]) {
            ans = false;
            break;
        }
    }

    ATYN(ans)

    return 0;
}