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`.

Examples

Input

azzel
apple


Output

Yes


Input

chokudai
redcoder


Output

No


Input

abcdefghijklmnopqrstuvwxyz
ibyhqfrekavclxjstdwgpzmonu


Output

Yes
import java.util.*;

import static java.lang.System.*;

public class Main {
    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        String S = ns();
        String T = ns();

        Map<Character, Character> s2T = new HashMap<>();
        Map<Character, Character> t2S = new HashMap<>();
        for (int i = 0; i < S.length(); i++) {
            char s = S.charAt(i);
            char t = T.charAt(i);

            if (t2S.containsKey(t) && t2S.get(t) != s) {
                out.println("No");
                return;
            }

            if (s2T.containsKey(s) && s2T.get(s) != t) {
                out.println("No");
                return;
            }

            s2T.put(s, t);
            t2S.put(t, s);
        }

        out.println("Yes");
    }

    static int ni() {
        return sc.nextInt();
    }

    static long nl() {
        return sc.nextLong();
    }

    static String ns() {
        return sc.next();
    }

    static int[] niarr(int N) {
        int[] a = new int[N];
        for (int i = 0; i < N; i++) {
            a[i] = sc.nextInt();
        }
        return a;
    }

    static long[] nlarr(int N) {
        long[] a = new long[N];
        for (int i = 0; i < N; i++) {
            a[i] = sc.nextLong();
        }
        return a;
    }
}