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;
const double eps(1e-8);
int n, m;
int f[1010][1010][2][2];
int ans[1010], g[1010], c[1010][1010];
void Add(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
void Mul(int &a, int b) { a = ((long long)a * (long long)b) % 1000000007; }
void Preit() {
  g[0] = 1;
  for (int i = 1; i <= 1000; i++) {
    g[i] = g[i - 1];
    Mul(g[i], i);
  }
  for (int i = 0; i <= 1000; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      Add(c[i][j], c[i - 1][j - 1]), Add(c[i][j], c[i - 1][j]);
  }
}
int main() {
  Preit();
  scanf("%d%d", &n, &m);
  memset(f, 0, sizeof(f));
  f[1][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= n; j++)
      for (int a = 0; a <= 1; a++)
        for (int b = 0; b <= 1; b++) {
          if (f[i][j][a][b] == 0) continue;
          Add(f[i + 1][j][b][0], f[i][j][a][b]);
          if (a == 0) Add(f[i + 1][j + 1][b][0], f[i][j][a][b]);
          if (i != n) Add(f[i + 1][j + 1][b][1], f[i][j][a][b]);
        }
  for (int i = n; i >= m; i--) {
    ans[i] = 0;
    for (int a = 0; a <= 1; a++)
      for (int b = 0; b <= 1; b++) Add(ans[i], f[1 + n][i][a][b]);
    Mul(ans[i], g[n - i]);
    int tmp;
    for (int j = i + 1; j <= n; j++) {
      tmp = ans[j];
      Mul(tmp, c[j][i]);
      ans[i] = (ans[i] - tmp + 1000000007) % 1000000007;
    }
  }
  cout << ans[m] << endl;
  return 0;
}
