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 long long M = 1e9 + 7;
const int mxN = 1000;
long long F[mxN + 1][mxN + 1][5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  F[1][0][4] = 1;
  for (int i = 2; i <= n; ++i) {
    for (int j = 0; j <= min(i, n); ++j) {
      F[i][j][0] =
          ((i - 3 - j) * F[i - 1][j][0] +
           (i - 2 - j) * (F[i - 1][j][1] + F[i - 1][j][2] + F[i - 1][j][4]) +
           (i - 1 - j) * F[i - 1][j][3]) %
          M;
      if (j + 1 <= i - 1)
        F[i][j][0] =
            (F[i][j][0] + (j + 1) * (F[i - 1][j + 1][0] + F[i - 1][j + 1][4]) +
             j * (F[i - 1][j + 1][1] + F[i - 1][j + 1][2]) +
             (j - 1) * F[i - 1][j + 1][3]) %
            M;
      F[i][j][1] = (F[i - 1][j][2] + F[i - 1][j][3]) % M;
      if (j >= 1)
        F[i][j][1] = (F[i][j][1] + F[i - 1][j - 1][0] + F[i - 1][j - 1][1]) % M;
      F[i][j][2] = (F[i - 1][j][1] + F[i - 1][j][3]) % M;
      if (j >= 1)
        F[i][j][2] = (F[i][j][2] + F[i - 1][j - 1][0] + F[i - 1][j - 1][2]) % M;
      if (j >= 2) F[i][j][3] = F[i - 1][j - 2][4];
      F[i][j][4] = (F[i - 1][j][0] + F[i - 1][j][1] + F[i - 1][j][2] +
                    F[i - 1][j][3] + F[i - 1][j][4]) %
                   M;
    }
  }
  cout << (F[n][k][0] + F[n][k][1] + F[n][k][2] + F[n][k][3] + F[n][k][4]) % M;
}
