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 Maxn = 1005, p = 1e9 + 7;
int n, m;
long long ans, fac[Maxn], inv[Maxn], f[Maxn][Maxn][2][2];
long long get_inv(int x) {
  return x <= 1 ? 1 : (p - p / x) * get_inv(p % x) % p;
}
long long C(int x, int y) { return fac[x] * inv[y] % p * inv[x - y] % p; }
int main() {
  scanf("%d%d", &n, &m);
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++)
    fac[i] = fac[i - 1] * i % p, inv[i] = get_inv(fac[i]);
  f[1][0][0][0] = f[1][1][0][1] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j <= i; j++) {
      (f[i][j][0][0] += f[i - 1][j][0][0] + f[i - 1][j][1][0]) %= p;
      (f[i][j][1][0] += f[i - 1][j][0][1] + f[i - 1][j][1][1]) %= p;
      (f[i][j + 1][0][1] += f[i - 1][j][0][0] + f[i - 1][j][1][0]) %= p;
      (f[i][j + 1][1][1] += f[i - 1][j][0][1] + f[i - 1][j][1][1]) %= p;
      (f[i][j + 1][0][0] += f[i - 1][j][0][0]) %= p;
      (f[i][j + 1][1][0] += f[i - 1][j][0][1]) %= p;
    }
  for (int i = m; i <= n; i++)
    (ans += (((i - m) & 1) ? p - 1 : 1) * (f[n][i][0][0] + f[n][i][1][0]) % p *
            C(i, m) % p * fac[n - i]) %= p;
  printf("%lld", ans);
  return 0;
}
