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;
long long f[2050][2050][4], F[2050], fac[2050], inv[2050];
long long C(int n, int m) {
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
int main() {
  scanf("%d%d", &n, &m);
  fac[0] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++)
    inv[i] = inv[1000000007 % i] * (1000000007 - 1000000007 / i) % 1000000007;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % 1000000007;
    inv[i] = inv[i] * inv[i - 1] % 1000000007;
  }
  f[1][0][0] = f[1][1][1] = 1;
  for (int i = 2; i < n; i++) {
    f[i][0][0] = 1;
    for (int j = 1; j <= i; j++) {
      f[i][j][0] =
          (f[i - 1][j - 1][0] + f[i - 1][j][0] + f[i - 1][j][2]) % 1000000007;
      f[i][j][2] =
          (f[i - 1][j - 1][1] + f[i - 1][j][1] + f[i - 1][j][3]) % 1000000007;
      f[i][j][1] = (f[i - 1][j - 1][0] + f[i - 1][j - 1][2]) % 1000000007;
      f[i][j][3] = (f[i - 1][j - 1][1] + f[i - 1][j - 1][3]) % 1000000007;
    }
  }
  f[n][0][0] = 1;
  for (int j = 1; j <= n; j++) {
    f[n][j][0] =
        (f[n - 1][j - 1][0] + f[n - 1][j][0] + f[n - 1][j][2]) % 1000000007;
    f[n][j][2] =
        (f[n - 1][j - 1][1] + f[n - 1][j][1] + f[n - 1][j][3]) % 1000000007;
  }
  for (int i = 0; i <= n; i++)
    F[i] = (f[n][i][0] + f[n][i][2]) * fac[n - i] % 1000000007;
  long long ans = 0;
  for (int i = m; i <= n; i++) {
    long long sav = C(i, m) * F[i] % 1000000007;
    ans = (ans + ((i - m) & 1 ? 1000000007 - sav : sav)) % 1000000007;
  }
  printf("%I64d", ans);
  return 0;
}
