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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std; 

int main(){
	int H , W , K;
	long long int MOD = 1000000007;
	cin >> H >> W >> K;
	if(W==1){
		cout << 1 << endl;
		return 0;
	}                                                
	long long int dp[H+1][W];
	int fib[8] = {1,2,3,5,8,13,21,34};
	dp[0][0] = 1;
	for(int i=1;i<W;i++)dp[0][i] = 0;
	if(W==2){
		for(int i=1;i<H+1;i++){
			dp[i][0] = (dp[i-1][0] + dp[i-1][1])%MOD;
			dp[i][1] = dp[i][0]%MOD;
		}
		cout << dp[H][K-1] << endl;
		return 0;
	}
	for(int i=1;i<H+1;i++){
		dp[i][0] = (dp[i-1][0]*fib[W-2] + dp[i-1][1]*fib[W-3])%MOD;
		if(W==3){
			dp[i][1] = (dp[i-1][0] + dp[i-1][1] + dp[i-1][2])%MOD;
		}else{
			dp[i][1] = (dp[i-1][0]*fib[W-3] + dp[i-1][1]*fib[W-3] + dp[i-1][2]*fib[W-4])%MOD;
		}
	for(int j=2;j<W-2;j++){
			dp[i][j] = (dp[i-1][j-1]*fib[j-2]*fib[W-j-2] + dp[i-1][j]*fib[j-1]*fib[W-j-2] + dp[i-1][j+1]*fib[j-1]*fib[W-j-3])%MOD;
		}
		if(W==3){
			dp[i][2] = (dp[i-1][1] + dp[i-1][2]*fib[1])%MOD;
		}else{
			dp[i][W-2] = (dp[i-1][W-3]*fib[W-4] + dp[i-1][W-2]*fib[W-3] + dp[i-1][W-1]*fib[W-3])%MOD;
			dp[i][W-1] = (dp[i-1][W-2]*fib[W-3] + dp[i-1][W-1]*fib[W-2])%MOD;
		}
	}
	cout << dp[H][K-1] << endl;
	return 0;
}