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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
 
class Main{
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int h = fs.nextInt();
        int w = fs.nextInt();
        int k = fs.nextInt();
        long[][] dp = new long[h+1][w];
        int mod = 1000000007;
        dp[0][0] = 1;
        BitSet bitSet;
        for(int i=0;i<h;i++){
            for(int j=0;j<w;j++){
                for(int bit = 0;bit<1<<(w-1);bit++){
                    boolean ok = true;
                    for(int l=0;l<w-2;l++){
                        if((bit>>(l) & bit>>(l+1))==1){
                            ok = false;
                            break;
                        }
                    }
                    if(ok){
                        if(j>=1 && ((bit & (1<<(w-1-j))) > 0)){
                            dp[i+1][j-1] = (dp[i+1][j-1] + dp[i][j])%mod;
                        }
                        else if(j<w-1 && ((bit & (1<<(w-2-j))) > 0)){
                            dp[i+1][j+1] = (dp[i+1][j+1] + dp[i][j])%mod;
                        }
                        else{
                            dp[i+1][j] = (dp[i+1][j] + dp[i][j])%mod;
                        }
                    }
                }
            }
        }
        System.out.println(dp[h][k-1]);
    }
 
    static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;
 
        private boolean hasNextByte() {
            if (ptr < buflen) {
                return true;
            } else {
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (buflen <= 0) {
                    return false;
                }
            }
            return true;
        }
 
        private int readByte() {
            if (hasNextByte()) return buffer[ptr++];
            else return -1;
        }
 
        private boolean isPrintableChar(int c) {
            return 33 <= c && c <= 126;
        }
 
        public boolean hasNext() {
            while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
            return hasNextByte();
        }
 
        public String next() {
            if (!hasNext()) throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while (isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }
 
        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) {
                throw new NumberFormatException();
            }
            while (true) {
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                } else if (b == -1 || !isPrintableChar(b)) {
                    return minus ? -n : n;
                } else {
                    throw new NumberFormatException();
                }
                b = readByte();
            }
        }
 
        public int nextInt() {
            long nl = nextLong();
            if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
            return (int) nl;
        }
 
        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}