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, m, MO = 1e9 + 7;
long long f[1010][1010][2][2];
long long C[1010][1010];
long long jc[1010], ans[1010];
int main() {
  scanf("%d%d", &n, &m);
  f[0][0][1][0] = 1;
  C[0][0] = 1;
  jc[0] = 1;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= i; j++)
      if (i == j || i == 0 || j == 0)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MO;
  for (int i = 1; i <= n; i++) jc[i] = (jc[i - 1] * i) % MO;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++)
      for (int k = 0; k <= 1; k++)
        for (int w = 0; w <= 1; w++) {
          f[i][j][w][0] = (f[i][j][w][0] + f[i - 1][j][k][w]) % MO;
          if (k == 0)
            f[i][j + 1][w][0] = (f[i][j + 1][w][0] + f[i - 1][j][k][w]) % MO;
          if (i != n)
            f[i][j + 1][w][1] = (f[i][j + 1][w][1] + f[i - 1][j][k][w]) % MO;
        }
  for (int i = 0; i <= n; i++) {
    for (int k = 0; k <= 1; k++)
      for (int w = 0; w <= 1; w++) ans[i] = (ans[i] + f[n][i][k][w]) % MO;
    ans[i] = (ans[i] * jc[n - i]) % MO;
  }
  int sd = 0;
  for (int i = m + 1; i <= n; i++) {
    if (sd)
      ans[m] = (ans[m] + ans[i] * C[i][m] + MO) % MO;
    else
      ans[m] = (ans[m] - ans[i] * C[i][m] + MO) % MO;
    sd ^= 1;
  }
  printf("%lld\n", ans[m]);
  return 0;
}
