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 M = 1000000007;
const int MaxN = 1000;
const int MaxG = 1000;
inline int &modaddto(int &a, const int &b) {
  a = (a + b) % M;
  return a;
}
int main() {
  int n, nG;
  cin >> n >> nG;
  static int f[MaxN + 1][MaxG + 1][3][3];
  f[1][0][0][0] = 1;
  for (int i = 1; i < n; i++)
    for (int g = 0; g <= i; g++)
      for (int ta = 0; ta < 3; ta++)
        for (int tb = 0; tb < 3; tb++)
          if (f[i][g][ta][tb]) {
            int cur = f[i][g][ta][tb];
            modaddto(f[i + 1][g][0][0], cur);
            if (ta == 0 && tb == 0) {
              modaddto(f[i + 1][g + 2][1][1], cur);
              modaddto(f[i + 1][g][2][2], (long long)cur * (i - g - 1) % M);
              modaddto(f[i + 1][g - 1][2][2], (long long)cur * g % M);
            } else {
              if (ta == 1)
                modaddto(f[i + 1][g][2][1], cur);
              else
                modaddto(f[i + 1][g + 1][2][1], cur);
              if (tb == 1)
                modaddto(f[i + 1][g][1][2], cur);
              else
                modaddto(f[i + 1][g + 1][1][2], cur);
              modaddto(f[i + 1][g][2][2],
                       (long long)cur * (i - g - (ta != 1) - (tb != 1)) % M);
              modaddto(f[i + 1][g - 1][2][2],
                       (long long)cur * (g - (ta == 1) - (tb == 1)) % M);
            }
          }
  int res = 0;
  for (int ta = 0; ta < 3; ta++)
    for (int tb = 0; tb < 3; tb++) modaddto(res, f[n][nG][ta][tb]);
  cout << res << endl;
  return 0;
}
