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>
long long f[1001][1001][4];
long long c[1001][1001];
long long s[1001];
long long fact[1001];
int n, m;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k < 4; k++) {
        f[i][j][k] = 0;
      }
    }
  }
  fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = fact[i - 1] * i;
    fact[i] %= 1000000007;
  }
  f[0][0][0] = 1;
  f[0][1][2] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      f[i][j][0] = f[i - 1][j][0] + f[i - 1][j][1];
      if (j - 1 >= 0) f[i][j][0] += f[i - 1][j - 1][0];
      f[i][j][1] = f[i - 1][j][2] + f[i - 1][j][3];
      if (j - 1 >= 0) f[i][j][1] += f[i - 1][j - 1][2];
      if (j - 1 >= 0) f[i][j][2] = f[i - 1][j - 1][0] + f[i - 1][j - 1][1];
      if (j - 1 >= 0) f[i][j][3] = f[i - 1][j - 1][2] + f[i - 1][j - 1][3];
      f[i][j][0] %= 1000000007;
      f[i][j][1] %= 1000000007;
      f[i][j][2] %= 1000000007;
      f[i][j][3] %= 1000000007;
    }
  }
  c[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (j - 1 >= 0)
        c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      else
        c[i][j] = c[i - 1][j];
      c[i][j] %= 1000000007;
    }
  }
  for (int k = 0; k <= n; k++) {
    for (int i = 0; i < 2; i++) s[k] += f[n - 1][k][i];
    s[k] *= fact[n - k];
    s[k] %= 1000000007;
  }
  for (int k = n; k >= 0; k--) {
    for (int i = k + 1; i <= n; i++) {
      s[k] -= s[i] * c[i][k];
      s[k] %= 1000000007;
      if (s[k] < 0) s[k] += 1000000007;
    }
  }
  printf("%I64d\n", s[m]);
}
