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 maxn = 1002;
const int P = 1000000007, all = 3;
int a[maxn], c[maxn][maxn], f[maxn][maxn][all + 1];
int main() {
  int n, K, i, j, s, all = 3;
  scanf("%d%d", &n, &K);
  f[1][0][2] = 1;
  for (i = 1; i <= n; ++i)
    for (j = 0; j <= n; ++j)
      for (s = 0; s <= all; ++s) {
        if (!(s & 2)) (f[i + 1][j + 1][s * 2 & all] += f[i][j][s]) %= P;
        if (i + 1 <= n) (f[i + 1][j + 1][s * 2 + 1 & all] += f[i][j][s]) %= P;
        (f[i + 1][j][s * 2 & all] += f[i][j][s]) %= P;
      }
  for (i = 0; i <= n; ++i)
    for (s = 0; s <= all; ++s) a[i] = (a[i] + f[n + 1][i][s]) % P;
  for (s = 1, i = 1; i <= n; ++i)
    s = (long long)s * i % P, a[n - i] = (long long)a[n - i] * s % P;
  for (c[0][0] = 1, i = 1; i <= n; ++i) {
    c[i][0] = c[i][i] = 1;
    for (j = 1; j < i; ++j) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % P;
  }
  for (i = n; i >= 0; --i)
    for (j = i + 1; j <= n; ++j)
      a[i] = (a[i] - (long long)c[j][i] * a[j] % P + P) % P;
  printf("%d\n", a[K]);
  return 0;
}
