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;
long long ADDmod(long long a, long long b) { return (a + b) % 1000000007; }
int c[1001][1001] = {0};
long long res[1001] = {0};
long long f[1001][1001][4] = {0};
long long jie[1001];
int main() {
  int n, k;
  cin >> n >> k;
  jie[0] = 1;
  for (int i = 0; i <= n; i++) {
    c[i][0] = 1;
    if (i != 0) jie[i] = jie[i - 1] * i % 1000000007;
    for (int j = 1; j <= i; j++) c[i][j] = ADDmod(c[i - 1][j], c[i - 1][j - 1]);
  }
  f[0][0][2] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++)
      for (int k = 0; k < 4; k++) {
        if (k & 1)
          f[i][j + 1][k / 2 + 2] =
              ADDmod(f[i][j + 1][k / 2 + 2], f[i - 1][j][k]);
        if (i < n)
          f[i][j + 1][k / 2] = ADDmod(f[i][j + 1][k / 2], f[i - 1][j][k]);
        f[i][j][k / 2 + 2] = ADDmod(f[i][j][k / 2 + 2], f[i - 1][j][k]);
      }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 4; j++) res[i] = ADDmod(res[i], f[n][i][j]);
    res[i] = res[i] * jie[n - i] % 1000000007;
  }
  for (int i = n - 1; i >= 0; i--)
    for (int j = i + 1; j <= n; j++)
      res[i] = ADDmod(res[i], 1000000007 - res[j] * c[j][i] % 1000000007);
  cout << res[k] << endl;
  return 0;
}
