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 Mn = 1000 + 10;
const int M = 1e9 + 7;
long long f[Mn][Mn][2][3];
int n, k;
int main() {
  scanf("%d%d", &n, &k);
  f[1][0][0][0] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= n; j++) {
      {
        (f[i + 1][j][0][0] += f[i][j][0][0]) %= M;
        (f[i + 1][j + 2][1][1] += f[i][j][0][0]) %= M;
        (f[i + 1][j][0][2] += (long long)(i - 1 - j) * f[i][j][0][0] % M) %= M;
        (f[i + 1][j - 1][0][2] += (long long)j * f[i][j][0][0] % M) %= M;
      }
      {
        (f[i + 1][j][0][0] += f[i][j][0][1]) %= M;
        (f[i + 1][j + 1][0][1] += f[i][j][0][1]) %= M;
        (f[i + 1][j][1][2] += f[i][j][0][1]) %= M;
        (f[i + 1][j][0][2] +=
         (long long)(i - 2 - (j - 1)) * f[i][j][0][1] % M) %= M;
        (f[i + 1][j - 1][0][2] += (long long)(j - 1) * f[i][j][0][1] % M) %= M;
      }
      {
        (f[i + 1][j][0][0] += f[i][j][0][2]) %= M;
        (f[i + 1][j + 1][0][1] += f[i][j][0][2]) %= M;
        (f[i + 1][j + 1][1][2] += f[i][j][0][2]) %= M;
        (f[i + 1][j][0][2] += (long long)(i - 2 - j) * f[i][j][0][2] % M) %= M;
        (f[i + 1][j - 1][0][2] += (long long)j * f[i][j][0][2] % M) %= M;
      }
      {
        (f[i + 1][j][0][0] += f[i][j][1][1]) %= M;
        (f[i + 1][j][0][1] += f[i][j][1][1]) %= M;
        (f[i + 1][j][1][2] += f[i][j][1][1]) %= M;
        (f[i + 1][j][0][2] +=
         (long long)(i - 2 - (j - 2)) * f[i][j][1][1] % M) %= M;
        (f[i + 1][j - 1][0][2] += (long long)(j - 2) * f[i][j][1][1] % M) %= M;
      }
      {
        (f[i + 1][j][0][0] += f[i][j][1][2]) %= M;
        (f[i + 1][j][0][1] += f[i][j][1][2]) %= M;
        (f[i + 1][j + 1][1][2] += f[i][j][1][2]) %= M;
        (f[i + 1][j][0][2] +=
         (long long)(i - 2 - (j - 1)) * f[i][j][1][2] % M) %= M;
        (f[i + 1][j - 1][0][2] += (long long)(j - 1) * f[i][j][1][2] % M) %= M;
      }
    }
  cout << (f[n][k][0][0] + f[n][k][0][1] + f[n][k][0][2] + f[n][k][1][1] +
           f[n][k][1][2]) %
              M
       << endl;
  return 0;
}
