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, ZZQ = 1e9 + 7;
int n, m, fac[N], C[N][N], f[N][N][2][2][2], ans;
int main() {
  int i, j, op1, op2, op3;
  cin >> n >> m;
  fac[0] = 1;
  for (i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % ZZQ;
  for (i = 0; i <= n; i++) C[i][0] = 1;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= i; j++) C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % ZZQ;
  f[0][0][0][0][0] = 1;
  for (i = 0; i <= n - 1; i++)
    for (j = 0; j <= i; j++) {
      for (op1 = 0; op1 <= 1; op1++)
        for (op2 = 0; op2 <= 1; op2++)
          for (op3 = 0; op3 <= 1; op3++)
            (f[i + 1][j][op2][op3][0] += f[i][j][op1][op2][op3]) %= ZZQ;
      if (i > 0)
        for (op1 = 0; op1 <= 1; op1++)
          for (op3 = 0; op3 <= 1; op3++)
            (f[i + 1][j + 1][1][op3][0] += f[i][j][op1][0][op3]) %= ZZQ;
      if (i < n - 1)
        for (op1 = 0; op1 <= 1; op1++)
          for (op2 = 0; op2 <= 1; op2++)
            for (op3 = 0; op3 <= 1; op3++)
              (f[i + 1][j + 1][op2][op3][1] += f[i][j][op1][op2][op3]) %= ZZQ;
    }
  for (i = m; i <= n; i++) {
    int tmp = 0;
    for (op1 = 0; op1 <= 1; op1++)
      for (op2 = 0; op2 <= 1; op2++) tmp = (tmp + f[n][i][op1][op2][0]) % ZZQ;
    int delta = 1ll * tmp * fac[n - i] % ZZQ * C[i][m] % ZZQ;
    if (i - m & 1)
      ans = (ans - delta + ZZQ) % ZZQ;
    else
      ans = (ans + delta) % ZZQ;
  }
  cout << ans << endl;
  return 0;
}
