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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

const int N = 10005;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, pii> node;

string s, t;
int mapeds[27], mapedt[27];
bool isRights[27], isWrongs[27];
bool isRightt[27], isWrongt[27];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
    memset(mapeds, -1, sizeof(mapeds));
    memset(mapedt, -1, sizeof(mapedt));
    cin >> s >> t;
    for (int i = 0; i < s.size(); i++)
    {
        if (s[i] == t[i])
        {
            if (isWrongs[s[i] - 'a'] || isWrongt[t[i] - 'a'])
            {
                cout << "No" << endl;
                return 0;
            }
            isRights[s[i] - 'a'] = true;
            isRightt[t[i] - 'a'] = true;
        }
        else
        {
            if ((isWrongs[s[i] - 'a'] || isRights[s[i] - 'a']) && (isWrongt[t[i] - 'a'] || isRightt[t[i] - 'a']))
            {
                if (isRights[s[i] - 'a'] || isRightt[t[i] - 'a'])
                {
                    cout << "No" << endl;
                    return 0;
                }
                if (mapeds[s[i] - 'a'] != -1 && mapeds[s[i] - 'a'] != mapedt[t[i] - 'a'])
                {
                    cout << "No" << endl;
                    return 0;
                }
            }

            // if(mapedt[t[i] - 'a'] != -1 && mapedt[t[i] - 'a'] != t[i] - 'a')
            // {
            //     cout << "No" << endl;
            //     return 0;
            // }

            mapeds[s[i] - 'a'] = t[i] - 'a';
            mapedt[t[i] - 'a'] = t[i] - 'a';
            isWrongs[s[i] - 'a'] = true;
            isWrongt[t[i] - 'a'] = true;
        }
    }
    cout << "Yes" << endl;
    return 0;
}