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;
int n, k;
long long mul[1005], c[1005][1005], f[1005][1005][2][3], chose[1005];
long long ans;
int main() {
  scanf("%d%d", &n, &k);
  mul[0] = 1;
  for (int i = 1; i <= n; i++) mul[i] = mul[i - 1] * (long long)i % 1000000007;
  for (int i = 0; i <= 1000; i++) c[i][0] = 1;
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= 1000; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  for (int i = 0; i <= n; i++) f[i][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i; j++) {
      f[i][j][0][0] = f[i - 1][j][0][0] + f[i - 1][j][0][1] +
                      f[i - 1][j][1][0] + f[i - 1][j][1][1];
      f[i][j][0][0] %= 1000000007;
      if (j) f[i][j][0][1] = f[i - 1][j - 1][0][0] + f[i - 1][j - 1][0][1];
      f[i][j][0][1] %= 1000000007;
      if (j)
        f[i][j][0][2] = f[i - 1][j - 1][0][0] + f[i - 1][j - 1][0][1] +
                        f[i - 1][j - 1][1][0] + f[i - 1][j - 1][1][1];
      f[i][j][0][2] %= 1000000007;
      f[i][j][1][0] = f[i - 1][j][0][2] + f[i - 1][j][1][2];
      f[i][j][1][0] %= 1000000007;
      if (j) f[i][j][1][1] = f[i - 1][j - 1][0][2];
      f[i][j][1][1] %= 1000000007;
      if (j) f[i][j][1][2] = f[i - 1][j - 1][0][2] + f[i - 1][j - 1][1][2];
      f[i][j][1][2] %= 1000000007;
    }
  for (int i = 0; i <= n; i++) {
    chose[i] = f[n][i][0][0] + f[n][i][0][1] + f[n][i][1][0] + f[n][i][1][1];
    chose[i] %= 1000000007;
  }
  for (int i = k, opt = 1; i <= n; i++, opt = -opt) {
    ans += chose[i] * mul[n - i] % 1000000007 * c[i][k] % 1000000007 * opt;
    ans = (ans % 1000000007 + 1000000007) % 1000000007;
  }
  printf("%I64d\n", ans);
}
