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: Accepted)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    char[] s, t;
    int[] as, bs;
    public static void main(String[] args) {
        Main m = new Main();
        m.read();
        m.solve();
    }

    private void read() {
        Scanner s = new Scanner(System.in);
        this.s = s.nextLine().toCharArray();
        this.t = s.nextLine().toCharArray();
        as = new int[26];
        bs = new int[26];
    }

    private void solve() {
        Arrays.fill(as, -1);
        Arrays.fill(bs, -1);

        boolean check = itr(s, as, t) && itr(t, bs, s);
        if (check)
            System.out.println("Yes");
        else
            System.out.println("No");
    }

    private boolean itr(char[] src, int[] trg, char[] cmp) {
        boolean res = true;
        for (int i = 0; i < src.length; i++) {
            int ix = src[i] - 'a';
            if (trg[ix] < 0)
                trg[ix] = cmp[i];
            else
                res &= trg[ix] == cmp[i];
        }
        return res;
    }
}