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 maxn = 1005, mod = 1e9 + 7;
int n, K;
long long f[maxn][maxn][6];
int main() {
  scanf("%d%d", &n, &K);
  f[3][0][0] = 1;
  f[3][2][1] = 1;
  f[3][0][2] = 1;
  f[3][2][2] = 1;
  f[3][2][3] = 1;
  f[3][2][4] = 1;
  for (int i = 3; i < n; i++) {
    for (int k = 0; k <= i; k++) {
      for (int j = 0; j < 5; j++)
        f[i + 1][k][2] = (f[i][k][j] + f[i + 1][k][2]) % mod;
      if (k < i - 1) {
        if (k)
          f[i + 1][k - 1][0] =
              (k * f[i][k][0] % mod + f[i + 1][k - 1][0]) % mod;
        f[i + 1][k][0] =
            ((i - 2 - k) * f[i][k][0] % mod + f[i + 1][k][0]) % mod;
        f[i + 1][k + 1][3] = (f[i][k][0] + f[i + 1][k + 1][3]) % mod;
        f[i + 1][k + 1][1] = (f[i][k][0] + f[i + 1][k + 1][1]) % mod;
      }
      if (k < i) {
        if (k)
          f[i + 1][k - 1][0] =
              ((k - 1) * f[i][k][1] % mod + f[i + 1][k - 1][0]) % mod;
        f[i + 1][k][0] =
            ((i - 1 - k) * f[i][k][1] % mod + f[i + 1][k][0]) % mod;
        f[i + 1][k][3] = (f[i][k][1] + f[i + 1][k][3]) % mod;
        f[i + 1][k + 1][1] = (f[i][k][1] + f[i + 1][k + 1][1]) % mod;
        if (k)
          f[i + 1][k - 1][0] =
              (k * f[i][k][2] % mod + f[i + 1][k - 1][0]) % mod;
        f[i + 1][k][0] =
            ((i - 1 - k) * f[i][k][2] % mod + f[i + 1][k][0]) % mod;
        f[i + 1][k + 2][4] = (f[i][k][2] + f[i + 1][k + 2][4]) % mod;
        if (k) {
          f[i + 1][k - 1][0] =
              ((k - 1) * f[i][k][3] % mod + f[i + 1][k - 1][0]) % mod;
          f[i + 1][k][0] =
              ((i - k - 1) * f[i][k][3] % mod + f[i + 1][k][0]) % mod;
          f[i + 1][k + 1][3] = (f[i][k][3] + f[i + 1][k + 1][3]) % mod;
          f[i + 1][k][1] = (f[i][k][3] + f[i + 1][k][1]) % mod;
        }
      }
      if (k > 1) {
        f[i + 1][k - 1][0] =
            ((k - 2) * f[i][k][4] % mod + f[i + 1][k - 1][0]) % mod;
        f[i + 1][k][0] = ((i - k) * f[i][k][4] % mod + f[i + 1][k][0]) % mod;
        f[i + 1][k][3] = (f[i][k][4] + f[i + 1][k][3]) % mod;
        f[i + 1][k][1] = (f[i][k][4] + f[i + 1][k][1]) % mod;
      }
    }
  }
  if (n == 1) {
    if (K == 1)
      cout << 0;
    else
      cout << 1;
    return 0;
  } else if (n == 2) {
    if (K == 0)
      cout << 1;
    else if (K == 2)
      cout << 1;
    else
      cout << 0;
    return 0;
  }
  long long ans = 0;
  for (int j = 0; j < 5; j++) ans = (f[n][K][j] + ans) % mod;
  cout << ans;
  return 0;
}
