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)

// Template
#include <iostream>
#include <string>
#include <vector>
#include <utility>
#include <functional>
#include <numeric>
#include <list>
#include <set>
#include <map>
#include <algorithm>
#include <cmath>

using namespace std;
const int MOD = 1000000007;

#define v_exists(elem, v) find(v.begin(),v.end(),elem)!=v.end()
#define s_exists(elem, s) s.find(elem)!=s.end()
#define mod_mul(x, y) x*y%MOD
#define mod_add(x, y) (x+y)%MOD
#define mod_sub(x, y) (x-y)%MOD
#define INF_INT 2147483647

long gcd(long a, long b) {
    return b == 0 ? a : gcd(b, a % b);
}

long lcm( long m, long n ){
    if ( ( 0 == m ) || ( 0 == n ) )
        return 0;

    return ((m / gcd(m, n)) * n);
}


long nCr(long n, long r){
    if(n==r || r==0){
        return 1;
    } else {
        return mod_add(nCr(n-1,r),nCr(n-1,r-1));
    }
}


int main(){
    
    string S,T; cin>>S>>T;
    int n = S.length();

    vector<int> s(27, -1);
    vector<int> t(27, -1);

    bool ans=true;

    for(int i=0; i<n;i++){
        int s_char = (int)S[i]-(int)'a';
        int t_char = (int)T[i]-(int)'a';

        if(s[s_char] == -1){
            s[s_char] = t_char;
        } else if (s[s_char] != t_char){
            ans = false;
            break;
        }

        if(t[t_char] == -1){
            t[t_char] = s_char;
        } else if (t[t_char] != s_char){
            ans = false;
            break;
        }    
        
    }

    if(ans) cout << "Yes" <<endl;
    else cout << "No" <<endl;


    return 0;

}