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 n, k, dp[1005][1005][2][2], mod = 1e9 + 7, ans[1005], fac[1005],
                                      c[1005][1005];
void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
long long f(int x, int y, int bef, int cur) {
  if (y == 0 && x == n + 1 && cur == 0)
    return 1;
  else if (x == n + 1 || y < 0)
    return 0;
  long long &ret = dp[x][y][bef][cur];
  if (ret != -1) return ret;
  ret = 0;
  if (x > 1 && bef == 0) add(ret, f(x + 1, y - 1, cur, 0));
  add(ret, f(x + 1, y - 1, cur, 1));
  add(ret, f(x + 1, y, cur, 0));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  memset(dp, -1, sizeof dp);
  for (int i = 0; i <= n; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= i; ++j)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
  }
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = (fac[i - 1] * i) % mod;
  for (int i = n; i >= 0; --i) {
    ans[i] = (f(1, i, 0, 0) * fac[n - i]) % mod;
    for (int j = i + 1; j <= n; ++j) add(ans[i], mod - ans[j] * c[j][i] % mod);
  }
  cout << ans[k] << '\n';
}
