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 inf = INT_MAX / 2;
const int Maxn = 1000;
const int Mod = 1000000007;
long long dp[Maxn + 5][Maxn + 5][2][2], n, k;
long long C[Maxn + 5][Maxn + 5], ans[Maxn + 5];
inline void mod_add(long long &x, long long y) {
  x += y;
  if (x > Mod) x -= Mod;
}
int main() {
  cin >> n >> k;
  C[0][0] = 1;
  for (int(i) = (1); (i) <= (n); ++(i)) {
    C[i][0] = 1;
    for (int(j) = (1); (j) <= (i); ++(j))
      mod_add(C[i][j], C[i - 1][j]), mod_add(C[i][j], C[i - 1][j - 1]);
  }
  dp[0][0][1][0] = 1;
  for (int(i) = (1); (i) <= (n); ++(i))
    for (int(j) = (0); (j) <= (i - 1); ++(j))
      for (int(now) = (0); (now) <= (1); ++(now))
        for (int(nxt) = (0); (nxt) <= (1); ++(nxt)) {
          if (!now) mod_add(dp[i][j + 1][nxt][0], dp[i - 1][j][now][nxt]);
          if (i < n) mod_add(dp[i][j + 1][nxt][1], dp[i - 1][j][now][nxt]);
          mod_add(dp[i][j][nxt][0], dp[i - 1][j][now][nxt]);
        }
  for (int(i) = (0); (i) <= (n); ++(i)) {
    for (int(now) = (0); (now) <= (1); ++(now))
      for (int(nxt) = (0); (nxt) <= (1); ++(nxt))
        mod_add(ans[i], dp[n][i][now][nxt]);
    for (int(j) = (2); (j) <= (n - i); ++(j)) ans[i] = ans[i] * j % Mod;
  }
  int flag = -1;
  for (int(i) = (k + 1); (i) <= (n); ++(i)) {
    ans[k] = (ans[k] + flag * C[i][k] * ans[i]) % Mod;
    if (ans[k] < 0) ans[k] += Mod;
    flag *= -1;
  }
  cout << ans[k];
  return 0;
}
