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 const maxn = 1005, mod = 1e9 + 7;
int n, k;
long long f[maxn][maxn][3][3], c[maxn][maxn], fac[maxn], ans[maxn];
void init() {
  fac[0] = 1;
  for (int i = 1; i <= maxn; i++) fac[i] = (fac[i - 1] * i * 1ll) % mod;
  for (int i = 1; i <= n; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  init();
  f[0][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i - 1; j++)
      for (int k = 0; k <= 1; k++)
        for (int l = 0; l <= 1; l++) {
          (f[i][j][l][0] += f[i - 1][j][k][l]) %= mod;
          if (!k) (f[i][j + 1][l][0] += f[i - 1][j][k][l]) %= mod;
          if (i < n) (f[i][j + 1][l][1] += f[i - 1][j][k][l]) %= mod;
        }
  for (int j = 0; j <= n; j++) {
    for (int k = 0; k <= 1; k++)
      for (int l = 0; l <= 1; l++) (ans[j] += f[n][j][k][l]) %= mod;
    (ans[j] *= fac[n - j]) %= mod;
  }
  for (int i = k + 1, fl = -1; i <= n; i++) {
    ans[k] = (ans[k] + fl * ans[i] * c[i][k] % mod + mod) % mod;
    fl = -fl;
  }
  printf("%I64d\n", ans[k]);
  return 0;
}
