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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
using System.IO;
using System.Linq;
using System.Diagnostics;
using System.Globalization;
using static System.Console;
using static System.Math;

namespace abc110_c
{
    class Program
    {
        static void Main(string[] args)
        {
            var s = Console.ReadLine().ToCharArray();
            var t = Console.ReadLine().ToCharArray();

            var dic = new Dictionary<char,char>();
            for(int i=0; i<s.Length; i++){
                if(dic.ContainsKey(s[i])){
                    if(dic[s[i]] != t[i]){
                        WriteLine("No");
                        return;
                    }
                }else {
                    dic[s[i]] = t[i];
                }
            }

            var dic2 = new Dictionary<char,char>();
            for(int i=0; i<t.Length; i++){
                if(dic2.ContainsKey(t[i])){
                    if(dic2[t[i]] != s[i]){
                        WriteLine("No");
                        return;
                    }
                }else {
                    dic2[t[i]] = s[i];
                }
            }
            WriteLine("Yes");
        }
    }
}