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 int p = 1e9 + 7, N = 1010;
int n, m, w, C[N][N], f[N][N][3], g[N][N][3], fac[N];
int main() {
  scanf("%d%d", &n, &m);
  w = (n + 1) / 2;
  if (n == 1 && m == 0) return printf("1\n"), 0;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = i * 1ll * fac[i - 1] % p;
  for (int i = 0; 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]) % p;
  }
  f[1][1][1] = f[1][0][2] = 1;
  for (int i = 2; i <= w; i++)
    for (int j = 0; j <= i; j++) {
      f[i][j][2] = (f[i - 1][j][0] + f[i - 1][j][1] * 1ll + f[i - 1][j][2]) % p;
      if (j == 0) continue;
      f[i][j][0] = (f[i - 1][j - 1][0] + f[i - 1][j - 1][2]) % p;
      f[i][j][1] =
          (f[i - 1][j - 1][0] + f[i - 1][j - 1][1] * 1ll + f[i - 1][j - 1][2]) %
          p;
    }
  g[1][1][0] = g[1][1][1] = g[1][0][2] = 1;
  for (int i = 2; i <= w; i++)
    for (int j = 0; j <= i; j++) {
      g[i][j][2] = (g[i - 1][j][0] + g[i - 1][j][1] * 1ll + g[i - 1][j][2]) % p;
      if (j == 0) continue;
      g[i][j][0] = (g[i - 1][j - 1][0] + g[i - 1][j - 1][2]) % p;
      g[i][j][1] =
          (g[i - 1][j - 1][0] + g[i - 1][j - 1][1] * 1ll + g[i - 1][j - 1][2]) %
          p;
    }
  int ans = 0;
  for (int i = m; i <= n; i++) {
    int sum = 0;
    for (int j = 0; j <= i; j++) {
      sum = (sum + (f[w][j][0] + f[w][j][2] * 1ll + (n % 2 ? 0 : f[w][j][1])) %
                       p *
                       (n % 2 ? (g[n - w][i - j][0] + g[n - w][i - j][2] * 1ll +
                                 g[n - w][i - j][1]) %
                                    p
                              : (f[w][i - j][0] + f[w][i - j][2] * 1ll +
                                 f[w][i - j][1]) %
                                    p)) %
            p;
    }
    ans = (ans + ((i - m) % 2 ? p - 1 : 1) * 1ll * C[i][m] % p * sum % p *
                     fac[n - i]) %
          p;
  }
  printf("%d\n", ans);
}
