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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll MOD = 1000000007;

int main() {
  int H, W, K;
  cin >> H >> W >> K;
  vector<vector<ll>> com(W, vector<ll>(W));
  for (int bit = 0; bit < (1 << (W - 1)); bit++) {
    int prev = 0;
    bool valid = true;
    for (int i = 0; i < W - 1; i++) {
      if (prev == 1 && ((bit >> i) & 1)) valid = false;
      prev = ((bit >> i) & 1);
    }
    if (!valid) continue;

    for (int i = 0; i < W; i++) {
      int left = ((bit >> (i - 1)) & 1) ? i > 0 : 0;
      int right = ((bit >> i) & 1) ? i < W - 1 : 0;
      if (left == 0 && right == 0) com[i][i]++;
      if (left == 0 && right == 1) com[i][i + 1]++, com[i + 1][i]++;
      // if (left == 1 && right == 0) com[i][i - 1]++, com[i - 1][i]++;
    }
  }

  vector<vector<ll>> dp(H + 1, vector<ll>(W, 0));
  dp[0][0] = 1;
  for (int h = 1; h <= H; h++) {
    for (int w = 0; w < W; w++) {
      if (w < W - 1) {
        dp[h][w] += (dp[h - 1][w + 1] * com[w][w + 1]) % MOD;
      }
      if (w > 0) {
        dp[h][w] += (dp[h - 1][w - 1] * com[w][w - 1]) % MOD;
      }
      dp[h][w] += (dp[h - 1][w] * com[w][w]) % MOD;
      dp[h][w] %= MOD;
    }
  }
  cout << dp[H][K - 1] << endl;
}