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>
int const mod = 1e9 + 7;
int n, k, fac[1010], dp[1010][4], ans;
int pow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1ll * res * x % mod;
    x = 1ll * x * x % mod, y >>= 1;
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &k);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  dp[0][2] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j; j--) {
      static int tmp[4];
      tmp[0] = (1ll * dp[j][0] + dp[j][2] + dp[j - 1][0]) % mod;
      tmp[1] = (dp[j - 1][0] + dp[j - 1][2]) % mod;
      tmp[2] = (1ll * dp[j][1] + dp[j][3] + dp[j - 1][1]) % mod;
      tmp[3] = (dp[j - 1][1] + dp[j - 1][3]) % mod;
      std::swap(tmp, dp[j]);
    }
    dp[0][2] = 0;
    dp[0][0] = 1;
  }
  for (int i = 0; i <= n; i++)
    dp[i][0] = 1ll * fac[n - i] * (dp[i][0] + dp[i][2]) % mod;
  for (int i = k; i <= n; i++)
    ans =
        (ans + ((i - k) & 1 ? mod - 1ll : 1ll) * dp[i][0] % mod * fac[i] % mod *
                   pow(fac[k], mod - 2) % mod * pow(fac[i - k], mod - 2)) %
        mod;
  printf("%d\n", ans);
  return 0;
}
