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 dp[1010][1010][2][2], comb[1010][1010], ans[1010], fact[1010];
int n, m;
int Dp() {
  comb[0][0] = 1, fact[0] = 1;
  for (int i = 1; i <= n; ++i) {
    comb[i][0] = 1, fact[i] = fact[i - 1] * i % 1000000007;
    for (int j = 1; j <= i; ++j)
      comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % 1000000007;
  }
  dp[0][0][0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      for (int a = 0; a <= 1; ++a)
        for (int b = 0; b <= 1; ++b) {
          dp[i][j][b][0] = (dp[i][j][b][0] + dp[i - 1][j][a][b]) % 1000000007;
          if (j >= 1 && i < n)
            dp[i][j][b][1] =
                (dp[i][j][b][1] + dp[i - 1][j - 1][a][b]) % 1000000007;
          if (a == 0 && i > 1 && j >= 1)
            dp[i][j][b][0] =
                (dp[i][j][b][0] + dp[i - 1][j - 1][a][b]) % 1000000007;
        }
  for (int i = n; i >= 0; --i) {
    for (int j = 0; j <= 1; ++j)
      for (int k = 0; k <= 1; ++k) ans[i] += dp[n][i][j][k];
    ans[i] = ans[i] * fact[n - i] % 1000000007;
    for (int j = i + 1; j <= n; ++j)
      ans[i] =
          (ans[i] - ans[j] * comb[j][i] % 1000000007 + 1000000007) % 1000000007;
  }
  return ans[m];
}
int main() {
  cin >> n >> m;
  cout << Dp() << endl;
  return 0;
}
