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 Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::cmp;

// Input macro from https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
  (source = $s:expr, $($r:tt)*) => {
    let mut iter = $s.split_whitespace();
    input_inner!{iter, $($r)*}
  };
  ($($r:tt)*) => {
    let s = {
      use std::io::Read;
      let mut s = String::new();
      std::io::stdin().read_to_string(&mut s).unwrap();
      s
    };
    let mut iter = s.split_whitespace();
    input_inner!{iter, $($r)*}
  };
}

macro_rules! input_inner {
  ($iter:expr) => {};
  ($iter:expr, ) => {};

  ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
    let $var = read_value!($iter, $t);
    input_inner!{$iter $($r)*}
  };
}

macro_rules! read_value {
  ($iter:expr, ( $($t:tt),* )) => {
    ( $(read_value!($iter, $t)),* )
  };

  ($iter:expr, [ $t:tt ; $len:expr ]) => {
    (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
  };

  ($iter:expr, chars) => {
    read_value!($iter, String).chars().collect::<Vec<char>>()
  };

  ($iter:expr, usize1) => {
    read_value!($iter, usize) - 1
  };

  ($iter:expr, $t:ty) => {
    $iter.next().unwrap().parse::<$t>().expect("Parse error")
  };
}

const MOD: i64 = 1_000_000_007;

#[derive(Copy, Clone, Hash, Debug)]
struct ModInt {
  x: i64
}
impl ModInt {
  fn new(x: i64) -> Self {
    ModInt {
      x: x % MOD
    }
  }
  fn pow(self, e: i64) -> Self {
    if e == 0 {
      Self::new(1)
    } else if e % 2 == 1 {
      self.pow(e - 1) * self
    } else {
      let y = self.pow(e / 2);
      y * y
    }
  }
  #[allow(dead_code)]
  fn inv(self) -> Self {
    self.pow(MOD - 2)
  }
}
impl<T: Into<ModInt>> std::ops::Add<T> for ModInt {
  type Output = Self;
  fn add(self, other: T) -> Self {
    let other = other.into();
    let sum = self.x + other.x;
    ModInt {
      x: if sum >= MOD { sum - MOD } else { sum }
    }
  }
}
impl<T: Into<ModInt>> std::ops::Sub<T> for ModInt {
  type Output = Self;
  fn sub(self, other: T) -> Self {
    let other = other.into();
    let sum = self.x - other.x;
    ModInt {
      x: if sum < 0 { sum + MOD } else { sum }
    }
  }
}
impl<T: Into<ModInt>> std::ops::Mul<T> for ModInt {
  type Output = Self;
  fn mul(self, other: T) -> Self {
    let other = other.into();
    ModInt {
      x: self.x * other.x % MOD
    }
  }
}
impl<T: Into<ModInt>> std::ops::Div<T> for ModInt {
  type Output = Self;
  fn div(self, other: T) -> Self {
    let other = other.into();
    self * other.inv()
  }
}
impl From<i64> for ModInt {
  fn from(x: i64) -> Self {
    ModInt::new(x)
  }
}
impl std::fmt::Display for ModInt {
  fn fmt(&self, f: &mut std::fmt::Formatter) -> std::fmt::Result {
    self.x.fmt(f)
  }
}
impl<T: Into<ModInt>> std::ops::AddAssign<T> for ModInt {
  fn add_assign(&mut self, other: T) {
    *self = *self + other;
  }
}
impl<T: Into<ModInt>> std::ops::SubAssign<T> for ModInt {
  fn sub_assign(&mut self, other: T) {
    *self = *self - other;
  }
}
impl<T: Into<ModInt>> std::ops::MulAssign<T> for ModInt {
  fn mul_assign(&mut self, other: T) {
    *self = *self * other;
  }
}
impl<T: Into<ModInt>> std::ops::DivAssign<T> for ModInt {
  fn div_assign(&mut self, other: T) {
    *self = *self / other;
  }
}

fn valid_row(mut x: u32) -> bool {
  while x > 0 {
    if x & 1 != 0 && x & 2 != 0 {
      return false;
    }
    x >>= 1;
  }
  true
}

fn main() {
  input! {
    h: usize,
    w: usize,
    k: usize1,
  }
  let mut dp = vec![vec![ModInt::new(0); w]; h + 1];
  dp[0][0] = 1.into();
  for r in 0 .. h {
    for lines in 0 .. 1 << (w - 1) {
      if ! valid_row(lines) {
        continue;
      }
      if lines & 1 == 0 {
        dp[r+1][0] += dp[r][0];
      }
      for c in 0 .. w - 1 {
        if (lines >> c) & 1 != 0 {
          dp[r+1][c] += dp[r][c+1];
          dp[r+1][c+1] += dp[r][c];
        } else if (lines >> (c + 1)) & 1 == 0 {
          dp[r+1][c+1] += dp[r][c+1];
        }
      }
    }
  }

  println!("{}", dp[h][k]);
}