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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

namespace CSharpSample01
{
    
    

    class Program
    {
      
        static void Main(string[] args)
        {

            var dic_str01 = new Dictionary<char, int>();
            var dic_str02 = new Dictionary<char, int>();

            string str01 = Console.ReadLine();
            for(int i = 0; i < str01.Length; i++)
            {
                if (!dic_str01.ContainsKey(str01[i]))
                {
                    dic_str01[str01[i]] = 1;
                }
            }
            string str02 = Console.ReadLine();
            
            var dic_match = new Dictionary<char, int>();
            var dic_unmatch = new Dictionary<char, int>();
            

            if (str01.Length != str02.Length)
            {
                Console.WriteLine("No");
                return;
            }

            int n = str02.Length;

            for(int i = 0; i < n; i++)
            {
                if (!dic_str01.ContainsKey(str02[i]))
                {
                    Console.WriteLine("No");
                    return;
                }
            }

            for(int i = 0; i < n; i++)
            {
                if (str01[i] != str02[i])
                {
                    dic_unmatch[str01[i]] = 1;
                    dic_unmatch[str02[i]] = 1;
                }
                else
                {
                    dic_match[str01[i]] = 1;                   
                }
            }

            //Console.WriteLine("unmatch");
            foreach(var s in dic_unmatch)
            {
                
                if (dic_match.ContainsKey(s.Key))
                {
                    Console.WriteLine("No");
                    return;
                }
            }

            //Console.WriteLine("match");            

            Console.WriteLine("Yes");

        }
    }
}