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 <algorithm>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <numeric>
#include <climits>
#include <iterator>
#include <iomanip>
#include <stack>
#include <set>
#include <cstdio>
#include <bitset>
#include <functional>
#include <tuple>
using namespace std;
const constexpr int INF = 1e9;
typedef pair<int, int> P;

#define FOR(i, a, n) for (ll i = (ll)a; i<(ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)

typedef long long ll;
typedef vector<int> VI;
const constexpr ll MOD = 1e9 + 7;
const int BITSIZE = 100010;

struct Less {
	bool operator()(const pair<int, int>& x, const pair<int, int>& y) const {
		return x.second < y.second;
	}
};
struct More {
	bool operator()(const pair<int, int>& x, const pair<int, int>& y) const {
		return x.first > y.first;
	}
};
//1-indexed
int bit[100010];

int bit_query(int i) {
	int res = 0;
	for (; i >= 1; i -= i & -i) res = max(res, bit[i]);
	return res;
}

void bit_set(int x, int v) {
	for (; x <= BITSIZE; x += x & -x) {
		bit[x] = max(bit[x], v);
	}
}

ll pow_mod(ll a, ll b){
    ll sum =1;
    while(b>0){
        if(b&1){
            sum *= a%MOD;
        }
        a = a* a%MOD;
        b >>= 1;
    }
    return sum;
}

int main() {
    int a[26];
    REP(i, 26) a[i]=-1;
    string s, t; cin >> s >> t;
    int n=s.size();
    
    bool ok = true;
    for(int i=0; i<s.size(); ++i){
        if(a[t[i]-'a']==-1){
            a[t[i]-'a'] = (int) s[i]-'a';
        }else{
            if(a[t[i]-'a']!=s[i]-'a')
                ok = false;
        }
    }
    if(ok) cout << "Yes" << endl;
    else cout << "No" << endl;
	return 0;
}