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.Generic;
using System.Linq;
using System.IO;
using static System.Console;
using static System.Int32;
using static System.Math;

class Program
{
    static void Main(string[] args)
    {
        label:
        WriteLine(hoge() ? "Yes" : "No");
        //goto label;
    }
    static bool hoge()
    {
        SortedList<char, List<int>> list1 = new SortedList<char, List<int>>(), list2 = new SortedList<char, List<int>>();
        var line = ReadLine();
        for (int i = 0; i < line.Length; ++i)
        {
            char c = line[i];
            if (!list1.ContainsKey(c)) list1.Add(c, new List<int>());
            list1[c].Add(i);
        }
        line = ReadLine();
        for (int i = 0; i < line.Length; ++i)
        {
            char c = line[i];
            if (!list2.ContainsKey(c)) list2.Add(c, new List<int>());
            list2[c].Add(i);
        }
        if (list1.Count != list2.Count) return false;
        var val1 = list1.Values.ToList();
        var val2 = list2.Values.ToList();
        return val1.TrueForAll(li1 => val2.Exists(li2 => piyo(li1, li2)));
    }

    static bool piyo(List<int> lhs, List<int> rhs)
    {
        if (lhs.Count != rhs.Count) return false;
        for (int i = 0; i < lhs.Count; ++i)
        {
            if (lhs[i] != rhs[i]) return false;
        }
        return true;
    }
}