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;
void add(int &x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
inline int read() {
  int x = 0;
  bool t = false;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') t = true, ch = getchar();
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - 48, ch = getchar();
  return t ? -x : x;
}
int n, m, jc[1010], a[1010], ans;
int f[1010][1010][4], C[1010][1010];
int main() {
  n = read();
  m = read();
  jc[0] = 1;
  for (int i = 1; i <= n; ++i) jc[i] = 1ll * jc[i - 1] * i % 1000000007;
  for (int i = 0; i <= n; ++i) C[i][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= i; ++j)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007;
  f[0][0][1] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < i; ++j)
      for (int S = 0; S < 4; ++S) {
        if (!f[i - 1][j][S]) continue;
        add(f[i][j][S >> 1], f[i - 1][j][S]);
        if (!(S & 1)) add(f[i][j + 1][S >> 1], f[i - 1][j][S]);
        if (i != n) add(f[i][j + 1][(S >> 1) | 2], f[i - 1][j][S]);
      }
  for (int i = 0; i <= n; ++i)
    for (int S = 0; S < 4; ++S)
      add(a[i], 1ll * f[n][i][S] * jc[n - i] % 1000000007);
  for (int i = m, d = 1; i <= n; ++i, d = 1000000007 - d)
    add(ans, 1ll * d * C[i][m] % 1000000007 * a[i] % 1000000007);
  printf("%d\n", ans);
  return 0;
}
