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)

// AtCoder Beginner Contest 110 - Problem C: String Transformation (https://abc110.contest.atcoder.jp/tasks/abc110_c)
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using ii = pair<ll, ll>;

#define REP(i, a, b) for (int (i) = (a); (i) <= (b); (i)++)
#define REPN(i, N) for (int (i) = 0; (i) < N; (i)++)
#define DEC(i, b, a) for (int (i) = (b); (i) >= (a); (i)--)
#define DECN(i, N) for (int (i) = N - 1; (i) >= 0; (i)--)

#define LOG(var, sep) (cerr << #var << " = " << (var) << (sep))
#define LOGM(msg) (cerr << (msg))
#define LOGV(vec, sep) { cerr << #vec << " ="; for (const auto& v : (vec)) cerr << " " << v; cerr << (sep); }
#define LOGA(arr, N, sep) { cerr << #arr << " ="; for (int i = 0; i < N; ++i) cerr << " " << (arr)[i]; cerr << (sep); }
#define LOGA2(arr, N, M, sep) { cerr << #arr << " =\n"; for (int i = 0; i < N; ++i) { for (int j = 0; j < M; ++j) cerr << (arr)[i][j] << " "; cerr << endl; } cerr << (sep); }

ostream& operator<<(ostream& os, const ii& p)
{
    os << "(" << p.first << ", " << p.second << ")";
    return os;
}

const int MAX { 2000010 };
const int oo { 2000000010 };
const int MOD { 1000000007 };

char histA[26], histB[26];

void solve(const string& S, const string& T)
{
    for (const auto& c : S)
        ++histA[c - 'a'];

    for (const auto& c : T)
        ++histB[c - 'a'];

    vector<int> xs, ys;

    REPN(i, 26)
    {
        if (histA[i])
            xs.push_back(histA[i]);

        if (histB[i])
            ys.push_back(histB[i]);
    }

    sort(xs.begin(), xs.end());
    sort(ys.begin(), ys.end());

    auto ans = (xs == ys ? "Yes" : "No");

    cout << ans << endl;
}

int main()
{
    ios::sync_with_stdio(false);

    string S, T;
    cin >> S >> T;

    solve(S, T);

    return 0;
}