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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"bufio"
	_ "container/heap"
	"fmt"
	"os"
	_ "sort"
	"strconv"
)

// I/O
type Scanner struct {
	sc *bufio.Scanner
}

func NewScanner() *Scanner {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, 1024), int(1e+9))
	return &Scanner{sc}
}

func (s *Scanner) nextStr() string {
	s.sc.Scan()
	return s.sc.Text()
}

func (s *Scanner) nextInt() int {
	i, e := strconv.Atoi(s.nextStr())
	if e != nil {
		panic(e)
	}
	return i
}

func (s *Scanner) nextFloat() float64 {
	f, e := strconv.ParseFloat(s.nextStr(), 64)
	if e != nil {
		panic(e)
	}
	return f
}

func (s *Scanner) nextRuneSlice() []rune {
	return []rune(s.nextStr())
}

func (s *Scanner) nextIntSlice(n int) []int {
	res := make([]int, n)
	for i := 0; i < n; i++ {
		res[i] = s.nextInt()
	}
	return res
}

func (s *Scanner) nextFloatSlice(n int) []float64 {
	res := make([]float64, n)
	for i := 0; i < n; i++ {
		res[i] = s.nextFloat()
	}
	return res
}

// Arithmetic
func max(nums ...int) int {
	m := nums[0]
	for _, i := range nums {
		if m < i {
			m = i
		}
	}
	return m
}

func min(nums ...int) int {
	m := nums[0]
	for _, i := range nums {
		if m > i {
			m = i
		}
	}
	return m
}

func abs(x int) int {
	if x > 0 {
		return x
	}
	return -x
}

func pow(x, y int) int {
	res := 1
	for i := 0; i < y; i++ {
		res *= x
	}
	return res
}

func ceil(a, b int) int {
	if a%b == 0 {
		return a / b
	} else {
		return a/b + 1
	}
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

func lcm(a, b int) int {
	return a / gcd(a, b) * b
}

// Sort
type RuneSlice []rune

func (a RuneSlice) Len() int           { return len(a) }
func (a RuneSlice) Less(i, j int) bool { return a[i] < a[j] }
func (a RuneSlice) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }

// Main
const MOD = int(1e+9) + 7
const INF = 1 << 60

type Point struct {
	h, w int
}

func main() {
	sc := NewScanner()
	wtr := bufio.NewWriter(os.Stdout)
	defer wtr.Flush()
	H, W, K := sc.nextInt(), sc.nextInt(), sc.nextInt()
	dp := make([][]int, H+1)
	for i := 0; i < H+1; i++ {
		dp[i] = make([]int, W)
	}
	dp[0][0] = 1
	for i := 0; i < H; i++ {
		for j := 0; j < W; j++ {
			var X, Y, Z int
			for bit := 0; bit < 1<<uint(W-1); bit++ {
				ok := true
				for k := 0; k < W-2; k++ {
					// 横棒が並んでいる場合
					if 1<<uint(k)&bit > 0 && 1<<uint(k+1)&bit > 0 {
						ok = false
					}
				}
				if !ok {
					continue
				}
				if j >= 1 && 1<<uint(j-1)&bit > 0 {
					// 左に進む
					X++
				} else if 1<<uint(j)&bit == 0 {
					// 横移動しない
					Y++
				} else if 1<<uint(j)&bit > 0 {
					// 右に進む
					Z++
				}
			}
			if j >= 1 {
				dp[i+1][j-1] += X * dp[i][j]
				dp[i+1][j-1] %= MOD
			}
			dp[i+1][j] += Y * dp[i][j]
			dp[i+1][j] %= MOD
			if j+1 < W {
				dp[i+1][j+1] += Z * dp[i][j]
				dp[i+1][j+1] %= MOD
			}
		}
	}
	fmt.Fprintln(wtr, dp[H][K-1])
}