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;
long long f[1013][1013][4];
long long p[1013];
long long c[1013][1013];
int N, K;
int main() {
  scanf("%d%d", &N, &K);
  f[0][0][1] = 1;
  for (int i = 0; i < N; ++i)
    for (int j = 0; j <= i; ++j) {
      for (int k = 0; k < 4; ++k) {
        f[i + 1][j + 1][(k >> 1) + 2] += f[i][j][k];
        f[i + 1][j + 1][(k >> 1) + 2] %= 1000000007LL;
        ;
        f[i + 1][j][(k >> 1)] += f[i][j][k];
        f[i + 1][j][(k >> 1)] %= 1000000007LL;
        ;
        if (!(k & 1)) {
          f[i + 1][j + 1][(k >> 1)] += f[i][j][k];
          f[i + 1][j + 1][(k >> 1)] %= 1000000007LL;
          ;
        }
      }
    }
  for (int i = 0; i <= N; ++i) {
    p[i] = f[N][i][0] + f[N][i][1];
    p[i] %= 1000000007LL;
    ;
    for (int j = N - i; j >= 1; --j) {
      p[i] = 1LL * p[i] * j;
      p[i] %= 1000000007LL;
      ;
    }
  }
  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 - 1] + c[i - 1][j]) % 1000000007LL;
  for (int i = N; i >= 0; --i)
    for (int j = i + 1; j <= N; ++j)
      (p[i] += 1000000007LL - p[j] * c[j][i] % 1000000007LL) %= 1000000007LL;
  cout << p[K];
  return 0;
}
