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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int pow2(int s){
    int t=1;
    for(int i=0;i<s;i++){
        t=t*2;
    }
    return t;
}
void hyouji(int d,int N,int x[][N]){
    for(int i=0;i<d;i++){
        for(int j=0;j<N;j++){
            printf("%d",x[i][j]);
        }
        printf("\n");
    }
}
long int move(int h,int w,int k,int W1,int Amida[][W1]){
    long int c=0;
    int x=0,y=1;
    int Now=Amida[y][x];
    for(int i=0;i<w-1;i++){
        if(i==0 && Amida[y][x+1]==1){
            x+=2;
            if(Amida[y][x-1]!=1 || Amida[y][x+1]!=1){
            y++;
            }else if(Amida[y][x-1]==1 && Amida[y][x+1]==1){
                return 0;
            }
        }else if(i==0 && Amida[y][x+1]!=1 ){
            y++;
        }else if(i!=0 && Amida[y][x+1]==1){
            x+=2;
            if(Amida[y][x-1]!=1 || Amida[y][x+1]!=1){
                y++;
            }else if(Amida[y][x-1]==1 && Amida[y][x+1]==1){
                return 0;
            }
        }else if(i!=0 && Amida[y][x-1]==1){
            x-=2;
            if(Amida[y][x-1]!=1 || Amida[y][x+1]!=1){
                y++;
            }else if(Amida[y][x-1]==1 && Amida[y][x+1]==1){
                return 0;
            }
        }else if(i!=0 && (Amida[y][x-1]!=1 && Amida[y][x+1]!=1)){
            y++;
        }else if(i!=0 && (Amida[y][x-1]==1 && Amida[y][x+1]==1)){
            return c;
        }else{}
        if(Amida[y][x-1]==1 && Amida[y][x+1]==1)return 0;
        if(y==h+1)break;
    }
                 if(x==(2*k-2))c++;
                 return c;
}

                 
int main(void){
    
    long int C=0,R;
    int H,K;
    int W;
    scanf("%d",&H);scanf("%d",&W);scanf("%d",&K);
    int W1=2*W-1;
    int Amida[H+2][W1];
    for(int i=0;i<H+2;i++){
        for(int j=0;j<W1;j++){
            Amida[i][j]=0;
        }
    }
    C=C+move(H,W,K,W1,Amida);
    for(long int m=2;m<=pow2((H)*(W-1));m++){
                    for(int i=1;i<(W1);i=i+2){
                        for(int j=1;j<=H;j++){
                            R=pow2(H*(i/2)+j-1);
                            if((m-1)%R==0){
                                if(Amida[j][i]==1){
                                    Amida[j][i]=0;
                                }else if(Amida[j][i]==0){
                                    Amida[j][i]=1;
                                }
                            }
                        }
                    }
        printf("%ld\n",m);
        hyouji(H+2,W1,Amida);
        C=C+move(H,W,K,W1,Amida);
        printf("%ld\n",C);
        printf("\n");
                }
                printf("%ld\n",C);
                return 0;
        }