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 fact[1005], c[1005][1005] = {0}, f[1005][1005][4] = {0}, an[1005] = {0}, n,
                K;
inline void upd(int &x, int y) { x = (x + y) % 1000000007; }
int main() {
  scanf("%d%d", &n, &K);
  fact[0] = c[0][0] = 1;
  for (int i = 1; i <= n; i++)
    c[i][0] = 1, fact[i] = 1ll * fact[i - 1] * i % 1000000007;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
  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) upd(f[i][j + 1][(k >> 1) | 2], f[i - 1][j][k]);
        if (i < n) upd(f[i][j + 1][k >> 1], f[i - 1][j][k]);
        upd(f[i][j][(k >> 1) | 2], f[i - 1][j][k]);
      }
  for (int i = 0; i <= n; i++) {
    for (int k = 0; k < 4; k++) upd(an[i], f[n][i][k]);
    an[i] = 1ll * an[i] * fact[n - i] % 1000000007;
  }
  for (int i = n - 1; ~i; i--)
    for (int j = i + 1; j <= n; j++)
      upd(an[i], -1ll * c[j][i] * an[j] % 1000000007);
  printf("%d\n", (an[K] % 1000000007 + 1000000007) % 1000000007);
  return 0;
}
