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>
using namespace std;
int main(){
    int a;
    string s,t,o,p;
    char c[2],d[2];
    cin>>s>>t;
    o=s;
    p=t;
    for (int i=0; i<s.size(); i++) {
        c[0]=s[i];
        d[0]=t[i];
        if (c[0]!=d[0]) {
            
        
        for (int j=0; j<s.size(); j++) {
            if(s[j]==c[0])s[j]=d[0];
        }
        }
    }
    if(s==t)a=0;
    else a=1;
    t=o;
    s=p;
    
    for (int i=0; i<s.size(); i++) {
        c[0]=s[i];
        d[0]=t[i];
        if (c[0]!=d[0]) {
            
            
            for (int j=0; j<s.size(); j++) {
                if(s[j]==c[0])s[j]=d[0];
            }
        }
    }
    
    if(s==t)a+=0;
    else a+=1;
    if (a==0)cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
}