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 N = 1010;
const int modi = 1000000007;
int n, m, f[N][N][4], G[N];
int C[N][N], str[N];
void push(int &a, int &b) {
  b += a;
  if (b >= modi) b -= modi;
}
int main() {
  int i, j, k;
  scanf("%d%d", &n, &m);
  for (str[0] = i = 1; i <= n; i++) str[i] = (long long)str[i - 1] * i % modi;
  for (i = 0; i <= n; i++) {
    C[i][0] = 1;
    for (j = 1; j <= i; j++) {
      C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
      if (C[i][j] >= modi) C[i][j] -= modi;
    }
  }
  f[0][0][0] = 1;
  for (i = 0; i < n; i++)
    for (j = 0; j <= n; j++)
      for (k = 0; k < 4; k++)
        if (f[i][j][k]) {
          if (i > 0 && !(k & 2))
            push(f[i][j][k], f[i + 1][j + 1][(k << 1) & 3]);
          if (i < n - 1) push(f[i][j][k], f[i + 1][j + 1][((k << 1) & 3) ^ 1]);
          push(f[i][j][k], f[i + 1][j][(k << 1) & 3]);
        }
  for (j = 0; j <= n; j++)
    for (k = 0; k < 4; k++) push(f[n][j][k], G[j]);
  for (i = 0; i <= n; i++) G[i] = (long long)G[i] * str[n - i] % modi;
  for (i = n - 1; i >= 0; i--)
    for (j = i + 1; j <= n; j++)
      G[i] = ((G[i] - (long long)C[j][i] * G[j] % modi) + modi) % modi;
  cout << G[m] << endl;
  return 0;
}
