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 fac[1001], inv[1001];
long long f[1001][1001][2][2], ans[1001];
long long C(int n, int m) {
  if (n < m) return 0;
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
int main() {
  scanf("%d%d", &n, &m);
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++)
    inv[i] = inv[1000000007 % i] * (1000000007 - 1000000007 / i) % 1000000007;
  for (int i = 2; i <= n; i++)
    fac[i] = fac[i - 1] * i % 1000000007,
    inv[i] = inv[i - 1] * inv[i] % 1000000007;
  f[0][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= n; j++)
      for (int k = 0; k < 2; k++)
        for (int l = 0; l < 2; l++) {
          (f[i][j][l][0] += f[i - 1][j][k][l]) %= 1000000007;
          if (!j) break;
          if (!k) (f[i][j][l][0] += f[i - 1][j - 1][k][l]) %= 1000000007;
          (f[i][j][l][1] += f[i - 1][j - 1][k][l]) %= 1000000007;
        }
  for (int i = 0; i <= n; i++)
    ans[i] =
        (f[n][i][0][0] + f[n][i][1][0]) % 1000000007 * fac[n - i] % 1000000007;
  for (int i = m, x = 1; i <= n; i++, x = -x)
    (ans[n + 1] += ans[i] * C(i, m) % 1000000007 * x + 1000000007) %=
        1000000007;
  printf("%lld\n", ans[n + 1]);
  return 0;
}
