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: Runtime Error)

import java.util.*;

public class Main {

  public static void main (String[] args) {

    Scanner sc = new Scanner(System.in);

    String s = sc.next();
    String t = sc.next();

    sc.close();

    String[] S = s.split("");
    String[] T = t.split("");

    HashMap<String, Integer> mapS = new HashMap<String, Integer>();
    HashMap<String, Integer> mapT = new HashMap<String, Integer>();

    for (int i = 0; i < S.length; i++) {
      if (mapS.containsKey(S[i])) {
        int v = mapS.get(S[i]);
        mapS.put(S[i],v+1);
      } else {
        mapS.put(S[i],1);
      }
    }

    for (int i = 0; i < T.length; i++) {
      if (mapT.containsKey(T[i])) {
        int v = mapT.get(T[i]);
        mapT.put(T[i],v+1);
      } else {
        mapT.put(T[i],1);
      }
    }

    int[] arrS = new int[26];
    int[] arrT = new int[26];

    for (Integer val : mapS.values()) {
      arrS[val]++;
    }

    for (Integer val : mapT.values()) {
      arrT[val]++;
    }

    String result = "Yes";

    for (int i = 0; i < 26; i++) {
      if (arrS[i] != arrT[i]){
        result = "No";
        break;
      }
    }

    System.out.println(result);

  }

}