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: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;

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

    set<char> used;
    vector<char> table('z' + 1);
    for (char i = 'a'; i <= 'z'; i++)
    {
        table[i] = i;
    }

    for (size_t i = 0; i < s.length(); i++)
    {
        if (table[s[i]] == t[i])
        {
            u += table[s[i]];
        }
        else
        {
            char c1 = table[s[i]];
            char c2 = t[i];
            if (used.find(c2) != used.end())
            {
                cout << "No" << endl;
                return 0;
            }

            if (table[table[c1]] != table[c1])
            {
                table[table[c1]] = c2;
            }
            else
            {
                table[c1] = c2;
            }

            if (table[table[c2]] != table[c2])
            {
                table[table[c2]] = c1;
            }
            else
            {
                table[c2] = c1;
            }
            used.insert(c2);
            u += c2;
        }
    }

    if (u.compare(t) == 0)
    {
        cout << "Yes" << endl;
    }
    else
    {
        cout << "No" << endl;
    }

    return 0;
}