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 = 1004, P = 1000000007;
int n, k, f[N][N][2][2], g[N];
long long fac[N], C[N][N];
int main() {
  int i, j;
  scanf("%d%d", &n, &k);
  fac[0] = 1;
  for (i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % P;
  C[0][0] = 1;
  for (i = 1; i <= n; i++) {
    C[i][0] = C[i][i] = 1;
    for (j = 1; j < i; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % P;
  }
  for (i = 0; i <= n; i++)
    for (int x = 0; x < 2; x++)
      for (int y = 0; y < 2; y++) f[0][i][x][y] = 0;
  f[0][0][0][0] = 1;
  for (i = 1; i <= n; i++)
    for (j = 0; j <= i; j++) {
      f[i][j][0][0] = (f[i - 1][j][0][0] + f[i - 1][j][1][0]) % P;
      if ((j) && (i > 1))
        f[i][j][0][0] = (f[i][j][0][0] + f[i - 1][j - 1][0][0]) % P;
      f[i][j][0][1] = f[i][j][1][0] = f[i][j][1][1] = 0;
      if (!j) continue;
      f[i][j][1][0] = ((f[i - 1][j][1][1] + f[i - 1][j][0][1]) % P +
                       f[i - 1][j - 1][0][1]) %
                      P;
      if (i == n) continue;
      f[i][j][0][1] = (f[i - 1][j - 1][0][0] + f[i - 1][j - 1][1][0]) % P;
      f[i][j][1][1] = (f[i - 1][j - 1][0][1] + f[i - 1][j - 1][1][1]) % P;
    }
  for (i = n; i >= k; i--) {
    g[i] = fac[n - i] * (f[n][i][0][0] + f[n][i][1][0]) % P;
    for (j = i + 1; j <= n; j++) g[i] = (g[i] - C[j][i] * g[j] % P + P) % P;
  }
  printf("%d\n", g[k]);
  return 0;
}
