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 mod = 1000000007;
long long N, K, dp[1010][1010][4];
int main() {
  cin >> N >> K;
  dp[1][0][3] = 1;
  long long i, j;
  for (i = 2; i <= N; i++)
    for (j = 0; j <= i; j++) {
      dp[i][j][0] = (max(i - 3 - j, 0LL) * dp[i - 1][j][0] +
                     (j + 1) * dp[i - 1][j + 1][0]) %
                    mod;
      dp[i][j][0] = (dp[i][j][0] + max(i - 2 - j, 0LL) * dp[i - 1][j][1] +
                     j * dp[i - 1][j + 1][1]) %
                    mod;
      dp[i][j][0] = (dp[i][j][0] + max(i - 1 - j, 0LL) * dp[i - 1][j][2] +
                     max(j - 1, 0LL) * dp[i - 1][j + 1][2]) %
                    mod;
      dp[i][j][0] = (dp[i][j][0] + max(0LL, i - 2 - j) * dp[i - 1][j][3] +
                     (j + 1) * dp[i - 1][j + 1][3]) %
                    mod;
      dp[i][j][1] = (dp[i - 1][j][1] + 2 * dp[i - 1][j][2]) % mod;
      if (j > 0)
        dp[i][j][1] =
            (dp[i][j][1] + 2 * dp[i - 1][j - 1][0] + dp[i - 1][j - 1][1]) % mod;
      if (j > 1) dp[i][j][2] = dp[i - 1][j - 2][3];
      dp[i][j][3] = (dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2] +
                     dp[i - 1][j][3]) %
                    mod;
    }
  cout << (dp[N][K][0] + dp[N][K][1] + dp[N][K][2] + dp[N][K][3]) % mod << "\n";
  return 0;
}
