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: Runtime Error)

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

class Dmondai
{
    static void Main()
    {
        const long Mod= 1000000007;
        var line=Console.ReadLine().Split(' ').Select(x => Int32.Parse(x)).ToArray();
        var H = line[0];
        var W = line[1];
        var K = line[2];

        //縦棒が一つしかない場合は1
        if (W == 1) { Console.WriteLine(1); return; }

        //ビット探索
        //横の線のあるなしを全列挙する
        var bitList = new List<string>();
        for (int i=0;i<(1<<(W-1));++i)
        {
            //二進数に変換、
            var value=Convert.ToString((i), 2);
            //空いている桁は0埋めする
            var bit =value.PadLeft(W-1,'0');

            //横棒が二つ続いた場合、処理をスキップさせるためのフラグ
            var isContinue = true;

            //横棒が二つ続くことはないので、その場合を除く
            for (int j=0;j<W-2;++j)
            {
                if (bit[j] == '1' && bit[j + 1] == '1')
                {
                    isContinue = false;
                    break;
                }
            }

            //横棒が二つ続かないパターン（正当なあみだくじ）
            if (isContinue)
            {
                bitList.Add(bit);
            }
        }

        //ビットりすと内にある数を保存（横棒のパターン）
        //１を引いて後ですぐ使えるようにする
        long bitCount = bitList.Count()-1;

        //動的計画法
        //各ポイントにおいて何通りのパターンがあるかを保存し、
        //それを利用しつつ下へ下る
        //保存する器を用意
        //H横棒の数、いま行でどこにいるか、0～H-1
        //W横棒の数、いま列でどこにいるか、0～W-1
        var ansArray = new long[H,W];
        //行のループ
        for (int i=0;i<=H;++i)
        {
            //列のループ
            for (int j=0;j<W;++j)
            {
                if (i == 0)
                {
                    //0,0地点は始点＝1
                    if (j == 0) ansArray[i, j] = 1;
                    else ansArray[i, j] = 0;
                }
                //横棒が0のとき
                else if (j==0)
                {
                    var temp = ansArray[i - 1, j] * bitCount % Mod;
                    temp = (temp + ansArray[i - 1, j + 1]) % Mod;
                    ansArray[i, j] = temp;

                }
                //横棒がW-1のとき（一番右端）
                else if (j == W - 1)
                {
                    var temp = ansArray[i - 1, j] * bitCount % Mod;
                    temp = (temp + ansArray[i - 1, j - 1]) % Mod;
                    ansArray[i, j] = temp;
                }
                //それ以外
                else
                {
                    var temp = (ansArray[i - 1, j] + ansArray[i - 1, j - 1] + ansArray[i - 1, j + 1]) % Mod ;
                    ansArray[i, j] = temp;
                }
            }
        }
        Console.WriteLine(ansArray[H-1,K-1]);

    }
}