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# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;
using System.Linq;
//Dp[h,x] h=移動前の縦、x=移動後の横　結果=場合の数

namespace D
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] In = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
            int h = In[0];//横線が引ける段
            int w = In[1];//本数
            int e = In[2];//終点
            var Dp = new long[h+1, w];
            //最初は左の棒(0)からスタートし、二本目(1)に移動する
            Dp[0, 0] = 1;
            if (w > 1)
            {
                for (int i = 1; i < w; i++)
                {
                    Dp[0, i] = 0;
                }
            }
            int mod=1000000007;
            //左の棒に行く、とどまる、右の棒
            for(int i = 0; i < h; i++)
            {
                for(int j = 0; j < w; j++)
                {
                    //ある高さの横線の引き方は2^(w-1)
                    for(int k=0; k < 1 << (w - 1); k++)
                    {
                        //２連続で横線はダメ
                        bool ok = true;
                        for(int l = 0; l < w - 2; l++)
                        {
                            if(((k>>l) & 1)==1 && (((k >> (l + 1)) & 1)==1))
                            {
                                ok = false;
                            }
                        }
                        if (!ok) continue;
                        //左、下に行く
                        if(j>=1 && (((k >> (j - 1)) & 1) == 1))
                        {
                            Dp[i + 1, j - 1] += Dp[i, j];
                            Dp[i + 1, j - 1] %= mod;
                        }
                        //右、下に行く
                        else if(j<=w-2 && (((k >> j) & 1) == 1))
                        {
                            Dp[i + 1, j + 1] += Dp[i, j];
                            Dp[i + 1, j + 1] %= mod;
                        }
                        //そのまま下に行く
                        else
                        {
                            Dp[i + 1, j] += Dp[i, j];
                            Dp[i + 1, j] %=mod;
                        }
                    }
                }
            }
            Console.WriteLine(Dp[h, e - 1]);
        }
    }
}