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;
long long jc[1007], C[1007][1007], f[1007][1007][4], g[1007];
int main() {
  scanf("%d%d", &n, &k);
  jc[0] = 1;
  for (int i = 1; i <= n; ++i) jc[i] = jc[i - 1] * i % 1000000007ll;
  for (int i = 0; i <= n; ++i) C[i][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007ll;
  f[0][0][0] = 1;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= n; ++j) {
      if (j < n) {
        if (i)
          f[i + 1][j + 1][0] = (f[i + 1][j + 1][0] + f[i][j][0]) % 1000000007ll;
        if (i + 1 < n)
          f[i + 1][j + 1][1] = (f[i + 1][j + 1][1] + f[i][j][0]) % 1000000007ll;
        if (i)
          f[i + 1][j + 1][2] = (f[i + 1][j + 1][2] + f[i][j][1]) % 1000000007ll;
        if (i + 1 < n)
          f[i + 1][j + 1][3] = (f[i + 1][j + 1][3] + f[i][j][1]) % 1000000007ll;
        if (i + 1 < n)
          f[i + 1][j + 1][1] = (f[i + 1][j + 1][1] + f[i][j][2]) % 1000000007ll;
        if (i + 1 < n)
          f[i + 1][j + 1][3] = (f[i + 1][j + 1][3] + f[i][j][3]) % 1000000007ll;
      }
      f[i + 1][j][0] = (f[i + 1][j][0] + f[i][j][0]) % 1000000007ll;
      f[i + 1][j][2] = (f[i + 1][j][2] + f[i][j][1]) % 1000000007ll;
      f[i + 1][j][0] = (f[i + 1][j][0] + f[i][j][2]) % 1000000007ll;
      f[i + 1][j][2] = (f[i + 1][j][2] + f[i][j][3]) % 1000000007ll;
    }
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j < 4; ++j) g[i] = (g[i] + f[n][i][j]) % 1000000007ll;
    g[i] = g[i] * jc[n - i] % 1000000007ll;
  }
  for (int i = n - 1; i >= k; --i)
    for (int j = i + 1; j <= n; ++j)
      g[i] =
          (g[i] - g[j] * C[j][i] % 1000000007ll + 1000000007ll) % 1000000007ll;
  printf("%lld", g[k]);
}
