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;
int f[1005][1005][4];
int fac[1005];
int comb[1005][1005];
inline void A(int &x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
}
int F[1005];
int init() {
  fac[0] = 1;
  for (int i = (1); i <= (n); i++) fac[i] = fac[i - 1] * 1ll * i % 1000000007;
  for (int i = (0); i <= (n); i++) {
    comb[i][0] = comb[i][i] = 1;
    for (int j = (1); j <= (i - 1); j++) {
      comb[i][j] = comb[i - 1][j];
      A(comb[i][j], comb[i - 1][j - 1]);
    }
  }
  f[1][0][0] = 1;
  f[1][1][2] = 1;
  for (int i = (1); i <= (n); i++)
    for (int j = (0); j <= (n); j++)
      for (int zt = (0); zt <= (3); zt++)
        if (f[i][j][zt]) {
          if (!(zt & 1)) A(f[i + 1][j + 1][zt / 2], f[i][j][zt]);
          if (i + 2 <= n) A(f[i + 1][j + 1][zt / 2 + 2], f[i][j][zt]);
          A(f[i + 1][j][zt / 2], f[i][j][zt]);
        }
  for (int i = (n); i >= (0); i--) {
    F[i] = (f[n][i][0] + f[n][i][1]) % 1000000007;
    F[i] = F[i] * 1ll * fac[n - i] % 1000000007;
    for (int j = (i + 1); j <= (n); j++)
      F[i] = (F[i] - F[j] * 1ll * comb[j][i] % 1000000007) % 1000000007;
  }
}
int k;
int main() {
  scanf("%d%d", &n, &k);
  if (k == n) {
    printf("%d\n", !(k & 1));
    return 0;
  }
  init();
  printf("%d\n", (F[k] + 1000000007) % 1000000007);
  return 0;
}
