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 Scala (2.11.7) (judge verdict: Accepted)

object Main {
  def main(args: Array[String]): Unit = {
    val s = new Main()
    s.solve()
    s.out.flush()
  }
}

class Main {
  import java.io._
  import java.util.StringTokenizer

  import scala.collection.mutable
  import scala.util.Sorting
  import math.{abs, max, min}
  import mutable.{ArrayBuffer, ListBuffer}
  import scala.reflect.ClassTag

  val MOD = 1000000007
  val out = new PrintWriter(System.out)


  val pow2 = Array.ofDim[Int](20)
  pow2(0) = 1
  rep(pow2.length - 1)(i => pow2(i + 1) = pow2(i) * 2)

  def solve(): Unit = {
    val H, W, K = ni()

    def comb(n: Int) = {
      n match {
        case 6 => 21
        case 5 => 13
        case 4 => 8
        case 3 => 5
        case 2 => 3
        case 1 => 2
        case 0 => 1
      }
    }

    val memo = Array.fill[Long](H + 1, W)(-1)
    def step(h: Int, w: Int): Long = {
      if (memo(h)(w) == -1) {
        val v = if (h == 0) {
          if (w == 0) 1 else 0
        } else {
          var res = 0L
          // 左から
          if (w > 0) {
            val l = max(0, w - 2)
            val r = max(0, W - 2 - w)
            res += step(h - 1, w - 1) * comb(l) * comb(r) % MOD
          }
          // 右から
          if (w < W - 1) {
            val l = max(0, w - 1)
            val r = max(0, W - 3 - w)
            res += step(h - 1, w + 1) * comb(l) * comb(r)  % MOD
          }
          // 上から
          {
            val l = max(0, w - 1)
            val r = max(0, W - 2 - w)
            res += step(h - 1, w) * comb(l) * comb(r)  % MOD
          }
          res % MOD
        }
        memo(h)(w) = v
      }
      memo(h)(w)
    }

    val ans = step(H, K - 1)
    out.println(ans)
  }


  class InputReader(val stream: InputStream) {
    private val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private var tokenizer: StringTokenizer = _

    def next(): String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = next().toInt
    def nextLong(): Long = next().toLong
    def nextChar(): Char = next().charAt(0)
  }
  val sc = new InputReader(System.in)
  def ni(): Int = sc.nextInt()
  def nl(): Long = sc.nextLong()
  def nc(): Char = sc.nextChar()
  def ns(): String = sc.next()
  def ns(n: Int): Array[Char] = ns().toCharArray
  def na(n: Int): Array[Int] = map(n)(_ => ni())
  def na2(n: Int, offset: Int = 0): (Array[Int], Array[Int]) = {
    val A1, A2 = Array.ofDim[Int](n)
    rep(n) { i =>
      A1(i) = ni() + offset
      A2(i) = ni() + offset
    }
    (A1, A2)
  }
  def nm(n: Int, m: Int): Array[Array[Int]] = {
    val A = Array.ofDim[Int](n, m)
    rep(n) { i =>
      rep(m) { j =>
        A(i)(j) = ni()
      }
    }
    A
  }
  def nal(n: Int): Array[Long] = map(n)(_ => nl())
  def nm_c(n: Int, m: Int): Array[Array[Char]] = map(n) (_ => ns(m))
  def rep(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = offset
    val N = n + offset
    while(i < N) { f(i); i += 1 }
  }
  def rep_r(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = n - 1 + offset
    while(i >= offset) { f(i); i -= 1 }
  }

  def map[@specialized A: ClassTag](n: Int)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    rep(n)(i => res(i) = f(i))
    res
  }

  def sumL(as: Array[Int]): Long = {
    var s = 0L
    rep(as.length)(i => s += as(i))
    s
  }

  def cumSum(as: Array[Int]) = {
    val cum = Array.ofDim[Int](as.length + 1)
    rep(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}