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;
long long js[1005], g[1005], f[1005][1005][2][2], ny[1005];
int n, m;
void add(long long &x, long long y) { x = (x + y) % 1000000007; }
long long C(int n, int m) {
  if (n < m) return 0;
  return js[n] * ny[m] % 1000000007 * ny[n - m] % 1000000007;
}
int main() {
  cin >> n >> m;
  f[0][0][1][0] = 1;
  js[0] = 1;
  for (int i = 1; i <= n; i++) js[i] = js[i - 1] * (long long)i % 1000000007;
  for (int i = 0; i <= n - 1; i++) {
    for (int j = 0; j <= i; j++) {
      add(f[i + 1][j][0][0], f[i][j][1][0] + f[i][j][0][0]);
      add(f[i + 1][j][1][0], f[i][j][0][1] + f[i][j][1][1]);
      add(f[i + 1][j + 1][0][0], f[i][j][0][0]);
      add(f[i + 1][j + 1][0][1], f[i][j][0][0] + f[i][j][1][0]);
      add(f[i + 1][j + 1][1][0], f[i][j][0][1]);
      add(f[i + 1][j + 1][1][1], f[i][j][0][1] + f[i][j][1][1]);
    }
  }
  for (int j = 0; j <= n; j++) {
    g[j] =
        (f[n][j][0][0] + f[n][j][1][0]) % 1000000007 * js[n - j] % 1000000007;
  }
  ny[0] = ny[1] = 1;
  for (int i = 2; i <= n; i++)
    ny[i] = (-ny[1000000007 % i] * (long long)(1000000007 / i) + 1000000007) %
            1000000007;
  for (int i = 1; i <= n; i++) ny[i] = ny[i - 1] * ny[i] % 1000000007;
  long long ans = 0;
  long long v = 1;
  for (int j = m; j <= n; j++) {
    ans = (ans + g[j] * C(j, m) % 1000000007 * v + 1000000007) % 1000000007;
    v = -v;
  }
  printf("%I64d\n", ans);
}
