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 md = 1000000007;
int f[2001][2001][3][3], n, k;
long long c[2001][2001], fac[2001], ans[2001];
int main() {
  scanf("%d%d", &n, &k);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * (long long)i) % md;
  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]) % md;
  }
  f[1][0][0][0] = 1;
  if (n > 1) f[1][1][0][1] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i; j++)
      for (int l = 0; l < 2; l++)
        for (int m = 0; m < 2; m++) {
          f[i + 1][j][m][0] = (f[i + 1][j][m][0] + f[i][j][l][m]) % md;
          if (i + 2 <= n)
            f[i + 1][j + 1][m][1] =
                (f[i + 1][j + 1][m][1] + f[i][j][l][m]) % md;
          if (!l)
            f[i + 1][j + 1][m][0] =
                (f[i + 1][j + 1][m][0] + f[i][j][l][m]) % md;
        }
  for (int i = 0; i <= n; i++) {
    ans[i] = ((f[n][i][0][1] + f[n][i][0][0]) % md +
              (f[n][i][1][0] + f[n][i][1][1]) % md) %
             md;
    ans[i] = ans[i] * fac[n - i] % md;
  }
  for (int i = n - 1; i >= k; i--)
    for (int j = i + 1; j <= n; j++) {
      ans[i] = (ans[i] - c[j][i] * ans[j] % md + md) % md;
    }
  cout << ans[k] << endl;
  return 0;
}
