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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <random>
#include <chrono>
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
#define INF 1000000000
#define mod 1000000007


int H, W, K;

bool check(int j){
    for(int k=0; k<W-1; k++){
        if(j>>k & 1 && j>>(k+1) & 1) return false;
    }
    return true;
}


int main() {
    cin >> H >> W >> K;
    vector<vector<int>> dp(H+1, vector<int>(W));
    dp[0][0] = 1;
    if(W==1){
        cout << 1 << endl;
        return 0;
    }
    
    for(int i=0; i<H; i++){
        for(int j=0; j<1<<(W-1); j++){
            if(!check(j)) continue;
            
            for(int k=0; k<W-1; k++){
                if(j>>k & 1){
                    dp[i+1][k] += dp[i][k+1];
                    dp[i+1][k+1] += dp[i][k];
                    dp[i+1][k] %= mod;
                    dp[i+1][k+1] %= mod;
                }else{
                    if(k==0){
                        dp[i+1][k] += dp[i][k];
                        dp[i+1][k] %= mod;
                    }else{
                        if(!(j>>(k-1) & 1)){
                            dp[i+1][k] += dp[i][k];
                            dp[i+1][k] %= mod;
                        }
                        if(k==W-2){
                            dp[i+1][k+1] += dp[i][k+1];
                            dp[i+1][k+1] %= mod;
                        }
                    }
                }
            }
        }
    }
    
    cout << dp[H][K-1] << endl;
    return 0;
}