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 f[1010][1010][2][2][2], n, m, g[1010], fec[1010], c[1010][1010];
int main() {
  scanf("%d%d", &n, &m);
  f[1][0][0][0][0] = 1;
  if (n > 1) f[1][1][0][0][1] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= i; j++)
      for (int A = 0; A <= 1; A++)
        for (int B = 0; B <= 1; B++)
          for (int C = 0; C <= 1; C++)
            if (f[i][j][A][B][C]) {
              f[i + 1][j][B][C][0] =
                  (f[i + 1][j][B][C][0] + f[i][j][A][B][C]) % 1000000007;
              if (!B)
                f[i + 1][j + 1][1][C][0] =
                    (f[i + 1][j + 1][1][C][0] + f[i][j][A][B][C]) % 1000000007;
              if (i + 1 < n)
                f[i + 1][j + 1][B][C][1] =
                    (f[i + 1][j + 1][B][C][1] + f[i][j][A][B][C]) % 1000000007;
            }
  for (int i = 0; i <= n; i++)
    for (int A = 0; A <= 1; A++)
      for (int B = 0; B <= 1; B++)
        for (int C = 0; C <= 1; C++)
          g[i] = (g[i] + f[n][i][A][B][C]) % 1000000007;
  c[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  }
  fec[0] = 1;
  for (int i = 1; i <= n; i++) fec[i] = fec[i - 1] * 1ll * i % 1000000007;
  for (int i = 0; i <= n; i++) g[i] = g[i] * 1ll * fec[n - i] % 1000000007;
  for (int i = n - 1; i >= m; i--)
    for (int j = i + 1; j <= n; j++)
      g[i] =
          (g[i] + 1000000007 - g[j] * 1ll * c[j][i] % 1000000007) % 1000000007;
  printf("%d\n", g[m]);
}
