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 N = 1005;
const int mo = 1000000007;
int C[N][N], f[N][N][2][2], g[N];
int n, k;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (0); i <= (int)n; i++) C[i][0] = 1;
  for (int i = (1); i <= (int)n; i++)
    for (int j = (1); j <= (int)i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mo;
  f[0][0][1][0] = 1;
  for (int i = (0); i <= (int)n - 1; i++)
    for (int j = (0); j <= (int)i; j++)
      for (int a = (0); a <= (int)1; a++)
        for (int b = (0); b <= (int)1; b++) {
          int v = f[i][j][a][b];
          if (!v) continue;
          (((f[i + 1][j][b][0]) += (v)) >= mo ? f[i + 1][j][b][0] -= mo : 233);
          (((f[i + 1][j + 1][b][1]) += (v)) >= mo ? f[i + 1][j + 1][b][1] -= mo
                                                  : 233);
          if (!a)
            (((f[i + 1][j + 1][b][0]) += (v)) >= mo
                 ? f[i + 1][j + 1][b][0] -= mo
                 : 233);
        }
  for (int i = (0); i <= (int)n; i++) {
    for (int a = (0); a <= (int)1; a++)
      (((g[i]) += (f[n][i][a][0])) >= mo ? g[i] -= mo : 233);
    for (int j = (1); j <= (int)n - i; j++) g[i] = 1ll * g[i] * j % mo;
  }
  for (int i = n; i >= k; i--)
    for (int j = (i + 1); j <= (int)n; j++)
      (((g[i]) += (mo - 1ll * g[j] * C[j][i] % mo)) >= mo ? g[i] -= mo : 233);
  printf("%d\n", g[k]);
}
