Problem:
Score: 400 points

Problem Statement

Amidakuji is a traditional method of lottery in Japan.

To make an amidakuji, we first draw W parallel vertical lines, and then draw horizontal lines that connect them. The length of each vertical line is H+1 [cm], and the endpoints of the horizontal lines must be at 1, 2, 3, ..., or H [cm] from the top of a vertical line.

A valid amidakuji is an amidakuji that satisfies the following conditions:

No two horizontal lines share an endpoint.

The two endpoints of each horizontal lines must be at the same height.

A horizontal line must connect adjacent vertical lines.

Find the number of the valid amidakuji that satisfy the following condition, modulo 1\ 000\ 000\ 007: if we trace the path from the top of the leftmost vertical line to the bottom, always following horizontal lines when we encounter them, we reach the bottom of the K-th vertical line from the left.

For example, in the following amidakuji, we will reach the bottom of the fourth vertical line from the left.

Constraints

H is an integer between 1 and 100 (inclusive).

W is an integer between 1 and 8 (inclusive).

K is an integer between 1 and W (inclusive).

Input

Input is given from Standard Input in the following format:

H W K

Output

Print the number of the amidakuji that satisfy the condition, modulo 1\ 000\ 000\ 007.

Sample Input 1

1 3 2

Sample Output 1

1

Only the following one amidakuji satisfies the condition:

Sample Input 2

1 3 1

Sample Output 2

2

Only the following two amidakuji satisfy the condition:

Sample Input 3

2 3 3

Sample Output 3

1

Only the following one amidakuji satisfies the condition:

Sample Input 4

2 3 1

Sample Output 4

5

Only the following five amidakuji satisfy the condition:

Sample Input 5

7 1 1

Sample Output 5

1

As there is only one vertical line, we cannot draw any horizontal lines. Thus, there is only one amidakuji that satisfies the condition: the amidakuji with no horizontal lines.

Sample Input 6

15 8 5

Sample Output 6

437760187

Be sure to print the answer modulo 1\ 000\ 000\ 007.

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

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
	static void Main()
	{
		long mod = 1000000007;
		string[] input = Console.ReadLine().Split(' ');
		int h = int.Parse(input[0]);
		int w = int.Parse(input[1]);
    int k = int.Parse(input[2]);
		long[,] dp = new long[h+1,w];//見た行数と何列目にいるか
    dp[0,0] = 1;

    if(w > 1)
    {
      for(int i = 1; i <= h; i++)
      {
        for(int j = 0; j < w; j++)
        {
          for(int x = 0; x < Math.Pow(2,w-1); x++)
          {
            List<int> rowMemo = BinaryToNum(w-1,x);
            bool rowContinue = false;
            for(int y = 1; y < w-1; y++)
            {
              if(rowMemo[y-1] + rowMemo[y] == 2) rowContinue = true;
            }
            if(rowContinue) continue;

            if(j == 0)
            {
              if(rowMemo[j] == 1) dp[i,j] += dp[i-1,j+1];
              else dp[i,j] += dp[i-1,j];
            }else if(j == w-1)
            {
              if(rowMemo[j-1] == 1) dp[i,j] += dp[i-1,j-1];
              else dp[i,j] += dp[i-1,j];
            }else
            {
              if(rowMemo[j] == 1) dp[i,j] += dp[i-1,j+1];
              else if(rowMemo[j-1] == 1) dp[i,j] += dp[i-1,j-1];
              else dp[i,j] += dp[i-1,j];
            }
            dp[i,j] %= mod;
          }
          //Console.WriteLine(i+" "+j+" "+dp[i,j]);
        }
      }
      Console.WriteLine(dp[h,k-1]);
    }else Console.WriteLine(1);
    
	}

  static List<int> FactorialToNum(int n, int num)
  {//n個の数を並べ替えた方法をリストに直す。
  //n=1でnum=0なら{0}。 n=2でnum=0なら{0,1}、num=1なら{1,0}
    List<int> answersNum = new List<int>();
    List<int> answers = new List<int>();
    List<int> answersRest = new List<int>();
    int rest = 1;
    for(int i = 2; i < n; i++)
    {
      rest *= i;
    }
    
    for(int i = n-1; i > 0; i--)
    {
      answersNum.Add(num/rest);
      num %= rest;
      rest /= i;
    }
    answersNum.Add(0);

    for(int i = 0; i < n; i++)
    {
      answersRest.Add(i);
    }
    for(int i = 0; i < n; i++)
    {
      answers.Add(answersRest[answersNum[i]]);
      answersRest.RemoveAt(answersNum[i]);
    }
    
    return answers;
  }

  static List<int> BinaryToNum(int n, int num)
  {//2進数でn桁の数numをリストに直す。
  //n=2でnum=0なら{0,0}、num=1なら{0,1}、num=2なら{1,0}。
    List<int> answers = new List<int>();
    int rest = 1;
    for(int i = 0; i < n-1; i++)
    {
      rest *= 2;
    }
    
    for(int i = 0; i < n; i++)
    {
      answers.Add(num/rest);
      num %= rest;
      rest /= 2;
    }
    
    return answers;
  }  
  
}