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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {
	static final int MOD = 1_000_000_007;
    public static long addMod(long a, long b){
    	long tmp = (a+b)%MOD;
    	return tmp>=0 ? tmp : tmp+MOD;
    }
    public static long subMod(long a, long b){
    	return addMod(a,-b);
    }
    public static long multMod(long a, long b){
    	long tmp = (a*b)%MOD;
    	return tmp>=0 ? tmp : tmp+MOD;
    }
    public static long powerMod(long a, long x){ //calculate a^x
        if(x<0)return 0;
        if(x==0)return 1;
        if(x%2==0) {
        	long half = powerMod(a,x/2);
        	return multMod(half,half);
        }
        return (a*powerMod(a,x-1)) % MOD;
    }
    public static long inverseMod(long a){
        return powerMod(a,MOD-2);
    }
    public static long divMod(long a, long b){
    	return multMod(a, inverseMod(b));
    }
    public static long[] makeFactorialArray(int size){
        long[] array = new long[size];
        array[0]=1;
        for(int i=1;i<size;i++){
        	array[i]=multMod(array[i-1],i);
        }
        return array;
    }
    public static long[] makeInversedArray(long[] original){
        long[] array = new long[original.length];
        for(int i=0;i<original.length;i++){
        	array[i] = inverseMod(original[i]);
        }
        return array;
    }
    public static long combination(int n, int r, long[] fact, long[] factInv){
        return multMod(fact[n], multMod(factInv[r],factInv[n-r]));
    }
    
    public static long access(long[] array, int index){
    	if(index<0 || index>=array.length) return 1;
    	else return array[index];
    }
    
    public static long solve(int H, int W, int K){
    	long[][] dp = new long[H+1][W];
    	Arrays.fill(dp[0], 0); dp[0][0]=1;
    	//System.out.println(Arrays.toString(dp[0]));
    	
    	long[] simpleBar = new long[]{1,2,3,5,8,13,21,34,55};
    	
    	for(int h=1;h<=H;h++){
    		for(int w=0;w<W;w++){
    			dp[h][w] += multMod(dp[h-1][w], multMod(access(simpleBar,w-1), access(simpleBar,W-w-2)));
    			dp[h][w] %= MOD;
    			if(w>0) {
    				dp[h][w-1] += multMod(dp[h-1][w], multMod(access(simpleBar,w-2), access(simpleBar,W-w-2)));
    			    dp[h][w-1] %= MOD;
    			}
    			if(w<W-1){
    			    dp[h][w+1] += multMod(dp[h-1][w], multMod(access(simpleBar,w-1), access(simpleBar,W-w-3)));
    			    dp[h][w] %= MOD;	
    			} 
    		}
    		//System.out.println(Arrays.toString(dp[h]));
    	}
    	return dp[H][K];
    }
    
    public static void main(String[] args) {
    	long[] fact = makeFactorialArray(200001);
        long[] factInv = makeInversedArray(fact);
        Scanner sc = new Scanner(System.in);
        int H = sc.nextInt();
        int W = sc.nextInt();
        int K = sc.nextInt()-1;
        System.out.println(solve(H,W,K));
        
    }
}