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: Wrong Answer)

use std::io::prelude::*;

fn main() {
    let stdin = std::io::stdin();
    let mut reader = std::io::BufReader::new(stdin.lock());

    let mut ss = Vec::new();

    for _ in 0..2 {
        let mut s = String::new();
        let _ = reader.read_line(&mut s);
        let s = s.as_bytes().to_vec();
        ss.push(s);
    }

    let mut p = [b' '; 26];
    let mut ans = true;

    for (c1, c2) in ss[0].iter().zip(ss[1].iter()) {
        if *c1 != *c2 {
            if p[(*c2 - b'a') as usize] != *c1 && p[(*c2 - b'a') as usize] != b' ' {
                ans = false;
                break;
            } else {
                p[(*c2 - b'a') as usize] = *c1;
            }
        }
    }

    println!("{}", if ans { "Yes" } else { "No" });
}