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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)

use std::collections::HashMap;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn readStr() -> String {
    let mut str = read::<String>();
    str
}

fn addMap2(k: char, v: char,m: &mut HashMap<char, char>) -> () {
    if((m).contains_key(&'1')) {
        return;
    }
    let d = ! m.contains_key(&k);
    let b = !d && *m.get(&k).unwrap() != v;
    if d {
        (m).insert(k, v);
    } else if b {
        (m).insert('1', '1');
    }
}

fn main() {
    let mut l1 = readStr();
    let mut l2 = readStr();
    let mut map1 = &mut HashMap::new();
    let mut map2 = &mut HashMap::new();

    for i in 0..l1.len() {
        addMap2(l1.chars().nth(i).unwrap(), l2.chars().nth(i).unwrap(), map1);
        addMap2(l2.chars().nth(i).unwrap(), l1.chars().nth(i).unwrap(), map2);
    }
    if !(map1).contains_key(&'1') {
        if !(map2).contains_key(&'1') {
            println!("Yes");
        } else{
            println!("No");
        }
    } else {
        println!("No");
    }
}