Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1, p2, ..., pn.

We'll call position i (1 ≤ i ≤ n) in permutation p1, p2, ..., pn good, if |p[i] - i| = 1. Count the number of permutations of size n with exactly k good positions. Print the answer modulo 1000000007 (109 + 7).

Input

The single line contains two space-separated integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ n).

Output

Print the number of permutations of length n with exactly k good positions modulo 1000000007 (109 + 7).

Examples

Input

1 0


Output

1


Input

2 1


Output

0


Input

3 2


Output

4


Input

4 1


Output

6


Input

7 4


Output

328

Note

The only permutation of size 1 has 0 good positions.

Permutation (1, 2) has 0 good positions, and permutation (2, 1) has 2 positions.

Permutations of size 3:

  1. (1, 2, 3) — 0 positions
  2. <image> — 2 positions
  3. <image> — 2 positions
  4. <image> — 2 positions
  5. <image> — 2 positions
  6. (3, 2, 1) — 0 positions
#include <bits/stdc++.h>
using namespace std;
long long dp[1010][1010][2][2], n, k;
long long c[1010][1010], ans[1010];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    c[i][0] = c[i][i] = 1LL;
    for (int j = 1; j <= i - 1; ++j) {
      c[i][j] = (c[i][j] + c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
    }
  }
  dp[0][0][1][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= i - 1; ++j)
      for (int now = 0; now < 2; ++now)
        for (int ne = 0; ne < 2; ++ne) {
          dp[i][j][ne][0] =
              (dp[i][j][ne][0] + dp[i - 1][j][now][ne]) % 1000000007;
          if (!now)
            dp[i][j + 1][ne][0] =
                (dp[i][j + 1][ne][0] + dp[i - 1][j][now][ne]) % 1000000007;
          if (i < n)
            dp[i][j + 1][ne][1] =
                (dp[i][j + 1][ne][1] + dp[i - 1][j][now][ne]) % 1000000007;
        }
  }
  for (int i = 0; i <= n; ++i) {
    for (int now = 0; now < 2; ++now)
      for (int ne = 0; ne < 2; ++ne)
        ans[i] = (ans[i] + dp[n][i][now][ne]) % 1000000007;
    for (int j = 2; j <= n - i; ++j) ans[i] = ans[i] * j % 1000000007;
  }
  int flag = -1;
  for (int i = k + 1; i <= n; ++i) {
    ans[k] = (ans[k] + flag * c[i][k] * ans[i] + 1000000007) % 1000000007;
    flag *= -1;
  }
  printf("%lld\n", ans[k]);
}
