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 f[1005][1005][2][2], n, m, g[1005], C[1005][1005], fu[1005], Ans, fac[1005];
inline int gi() {
  int x;
  scanf("%d", &x);
  return x;
}
inline void add(int &x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
int main() {
  n = gi();
  m = gi();
  f[1][0][0][0] = 1;
  f[1][1][0][1] = (n > 1) ? 1 : 0;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= n; j++)
      for (int p1 = 0; p1 <= 1; p1++)
        for (int p2 = 0; p2 <= 1; p2++)
          if (f[i][j][p1][p2]) {
            if (!p1) add(f[i + 1][j + 1][p2][0], f[i][j][p1][p2]);
            if (i + 1 != n) add(f[i + 1][j + 1][p2][1], f[i][j][p1][p2]);
            add(f[i + 1][j][p2][0], f[i][j][p1][p2]);
          }
  for (int j = 0; j <= n; j++)
    for (int p1 = 0; p1 <= 1; p1++)
      for (int p2 = 0; p2 <= 1; p2++) add(g[j], f[n][j][p1][p2]);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * i * fac[i - 1] % 1000000007;
  for (int i = 0; i <= n; i++) g[i] = 1ll * g[i] * fac[n - i] % 1000000007;
  C[0][0] = 1;
  for (int i = 1, j; i <= n; i++)
    for (j = C[i][0] = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 1000000007;
  for (int i = 0; i <= n; i++) fu[i] = (i & 1) ? 1000000007 - 1 : 1;
  for (int j = m; j <= n; j++)
    add(Ans, 1ll * fu[j - m] * g[j] % 1000000007 * C[j][m] % 1000000007);
  cout << Ans;
  return 0;
}
