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 <cstdio>
#include <cstring>
#include <cmath>
#include <utility>
#include <iostream>
#include <functional>
#include <bitset>
#include <algorithm>
#include <vector>
#include <forward_list>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <tuple>
#include <numeric>
#define rep(i, s, g) for ((i) = (s); (i) < (g); ++(i))
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
const ll MOD = 1e9 + 7;
const ll INF = (1ll << 60);

int main(void)
{
    string s, t;
    cin >> s >> t;

    map<char, ll> sc, tc;

    for (int i = 0; i < s.size(); i++)
    {
        sc[s[i]]++;
    }

    for (int i = 0; i < t.size(); i++)
    {
        tc[t[i]]++;
    }

    map<ll, ll> s1, t1;

    for (auto x : sc)
    {
        s1[x.second]++;
    }

    for (auto x : tc)
    {
        t1[x.second]++;
    }

    for (int i = 0; i <= s.size(); i++)
    {
        if (s1[i] != t1[i])
        {
            cout << "No" << endl;
            return 0;
        }
    }

    cout << "Yes" << endl;
}