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;
int n, k;
int f[1010][1010][2][2];
int g[1010];
int C[1010][1010];
int fac[1010];
void updata(int &x, int y) {
  x = (x + y % 1000000007) % 1000000007;
  return;
}
int main() {
  scanf("%d%d", &n, &k);
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % 1000000007;
  f[0][0][1][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) {
      updata(f[i][j][0][0],
             (f[i - 1][j][0][0] + f[i - 1][j][1][0]) % 1000000007);
      updata(f[i][j][1][0],
             (f[i - 1][j][0][1] + f[i - 1][j][1][1]) % 1000000007);
      updata(f[i][j + 1][0][0], f[i - 1][j][0][0]);
      updata(f[i][j + 1][1][0], f[i - 1][j][0][1]);
      updata(f[i][j + 1][0][1], f[i - 1][j][0][0] + f[i - 1][j][1][0]);
      updata(f[i][j + 1][1][1], f[i - 1][j][0][1] + f[i - 1][j][1][1]);
    }
  }
  for (int i = 0; i <= n; ++i) {
    g[i] = 1ll * (f[n][i][0][0] + f[n][i][1][0]) % 1000000007 * fac[n - i] %
           1000000007;
  }
  for (int i = 1; i <= n; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007;
    }
  }
  int ans = g[k];
  for (int p = k + 1; p <= n; ++p) {
    ans =
        (ans + (1ll * (((p - k) & 1) ? -1 : 1) * C[p][k] * g[p] % 1000000007) +
         1000000007) %
        1000000007;
  }
  cout << ans << endl;
  return 0;
}
