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;
using System.Collections.Generic;
using static System.Console;
using System.Text;
using System.IO;
using static System.Math;
using System.Numerics;

namespace AtCoder
{
    class Program
    {
        static public long[] Sarray() { return ReadLine().Split().Select(long.Parse).ToArray(); }
        static public List<long> Slist() { return ReadLine().Split().Select(long.Parse).ToList(); }
        static long Mod = (long)1e9 + 7;
        static void Main(string[] args)
        {
            SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
            Solve();
            Out.Flush();
            Read();
        }
        static void Solve()
        {
            var HWK = Sarray();
            var H = HWK[0];
            var W = HWK[1];
            var K = HWK[2];
            var free = new long[8];
            free[0] = 1;free[1] = 1;
            for (var i = 2; i < 8; ++i)
                free[i] = free[i - 1] + free[i - 2];
            var dp = new long[H + 2, W + 2];
            dp[1, 1] = 1;
            for (var h = 1; h <= H; ++h)
            {
                for (var w = 1; w <= W; ++w)
                {
                    var t = 0L;
                    if (1 != w)
                    {
                        t = free[w - 2] * free[W - w] % Mod;
                        t = t * dp[h, w] % Mod;
                        dp[h + 1, w - 1] = (t + dp[h + 1, w - 1]) % Mod;
                    }
                    t = free[w - 1] * free[W - w] % Mod;
                    t = t * dp[h, w] % Mod;
                    dp[h + 1, w] = (t + dp[h + 1, w]) % Mod;
                    if (W != w)
                    {
                        t = free[w - 1] * free[W - w - 1] % Mod;
                        t = t * dp[h, w] % Mod;
                        dp[h + 1, w + 1] = (t + dp[h + 1, w + 1]) % Mod;
                    }
                }
            }
            WriteLine(dp[H + 1, K]);
        }
    }
}