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, mod = 1000000007;
int n, m;
long long C[N][N], g[N][N][4], f[N], s[N];
int main() {
  scanf("%d%d", &n, &m);
  g[0][0][1] = 1;
  for (int i = 0; i <= n - 1; i++) {
    for (int j = 0; j <= n - 1; j++) {
      for (int k = 0; k <= 3; k++) {
        int w = g[i][j][k];
        if (!w) continue;
        if (k % 2 == 0) {
          (g[i + 1][j + 1][(k >> 1)] += w) %= mod;
        }
        if (i < n - 1) (g[i + 1][j + 1][(k >> 1) + 2] += w) %= mod;
        (g[i + 1][j][k >> 1] += w) %= mod;
      }
    }
  }
  s[0] = 1;
  for (int i = 1; i <= n; i++) {
    C[i][0] = C[0][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
    s[i] = s[i - 1] * i % mod;
  }
  for (int i = n; i >= m; i--) {
    for (int j = 0; j <= 3; j++) (f[i] += g[n][i][j]) %= mod;
    f[i] = f[i] * s[n - i] % mod;
    for (int j = i + 1; j <= n; j++)
      f[i] = (f[i] - (f[j] * C[j][i] % mod) + mod) % mod;
  }
  cout << f[m] << endl;
  return 0;
}
