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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.HashSet;
import java.util.Map;
import java.util.Set;
import java.util.concurrent.atomic.AtomicReference;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {
    private static java.util.Scanner scanner = new java.util.Scanner(System.in);

    static class C {
        char s, t;

        C(char s, char t) {
            this.s = s;
            this.t = t;
        }

        @Override
        public int hashCode() {
            return 0;
        }

        @Override
        public boolean equals(Object obj) {
            return obj instanceof C && ((C)obj).s == s ^ ((C)obj).t == t;
        }
    }

    public static void main(String[] args) {
        char[] s = scanner.next().toCharArray();
        char[] t = scanner.next().toCharArray();
        long size = IntStream.range(0, s.length)
                .boxed()
                .map(i -> new C(s[i], t[i]))
                .distinct()
                .count();
        System.out.println(s.length == size ? "Yes" : "No");
    }
}