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;
void upd(int &x, int v) {
  x += v;
  if (x >= 1000000007) x -= 1000000007;
}
int f[1005][1005][2][2];
int fact[1005];
int c[1005][1005];
int g[1005];
int h[1005];
int main() {
  scanf("%d%d", &n, &k);
  f[0][0][1][0] = 1;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = 1ll * fact[i - 1] * i % 1000000007;
  c[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= n; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      if (c[i][j] >= 1000000007) c[i][j] -= 1000000007;
    }
  }
  int v;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int p = 0; p < 2; p++)
        for (int q = 0; q < 2; q++)
          if (v = f[i][j][p][q]) {
            upd(f[i + 1][j][q][0], v);
            if (!p) upd(f[i + 1][j + 1][q][0], v);
            if (i + 1 != n) upd(f[i + 1][j + 1][q][1], v);
          }
  for (int i = 0; i <= n; i++) {
    g[i] =
        (0ll + f[n][i][0][0] + f[n][i][0][1] + f[n][i][1][0] + f[n][i][1][1]) %
        1000000007;
    g[i] = 1ll * g[i] * fact[n - i] % 1000000007;
  }
  h[n] = g[n];
  for (int i = n - 1; i >= k; i--) {
    h[i] = g[i];
    for (int t = i + 1; t <= n; t++) {
      upd(h[i], 1000000007 - 1ll * h[t] * c[t][i] % 1000000007);
    }
  }
  printf("%d\n", h[k] % 1000000007);
  return 0;
}
