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<vector>
#include<string>
#include<algorithm>

using namespace std;

int in_or_not(const vector<int> &c, int a){
  bool flag = false;
  int i;
  for(int i = 0; i < c.size(); ++i){
    if(c[i] == a){
      flag = true;
      break;
    }
  }
  if(flag){
    return i;
  }
  else{
    return -1;
  }
}

int main() {
  string a, b;
  cin >> a >> b;
  vector<int> ca(26);
  vector<int> cb(26);
  bool flag = true;
  int l = (int)a.size();

  for(int i = 0; i < l; ++i){
    int x = in_or_not(ca, a[i]);
    if(x == -1){
      ca.push_back(a[i]);
    }
    int y = in_or_not(cb, b[i]);
    if(y == -1){
      cb.push_back(b[i]);
    }
    if (x != y){
      flag = false;
      break;
    }
  }

  if(flag){
    cout << "Yes" << endl;
  }
  else{
    cout << "No" << endl;
  }
}