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 = 2100;
const int MOD = 1e9 + 7;
int n, k;
long long f[MAXN][MAXN][2][2], c[MAXN][MAXN], jc[MAXN], ans[MAXN];
int main() {
  scanf("%d%d", &n, &k);
  f[0][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i - 1; j++)
      for (int l = 0; l <= 1; l++)
        for (int t = 0; t <= 1; t++) {
          if (l == 0)
            f[i][j + 1][t][0] = (f[i][j + 1][t][0] + f[i - 1][j][l][t]) % MOD;
          if (i != n)
            f[i][j + 1][t][1] = (f[i][j + 1][t][1] + f[i - 1][j][l][t]) % MOD;
          f[i][j][t][0] = (f[i][j][t][0] + f[i - 1][j][l][t]) % MOD;
        }
  c[0][0] = jc[0] = 1;
  for (int i = 1; i <= n; i++) {
    c[i][0] = 1;
    jc[i] = ((long long)jc[i - 1] * i) % MOD;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= 1; j++)
      for (int l = 0; l <= 1; l++) ans[i] = (ans[i] + f[n][i][j][l]) % MOD;
    ans[i] = (ans[i] * jc[n - i]) % MOD;
  }
  for (int i = 1; i + k <= n; i++) {
    if (i & 1)
      ans[k] =
          ((long long)MOD + ans[k] - (c[i + k][k] * ans[i + k]) % MOD) % MOD;
    else
      ans[k] = (ans[k] + c[i + k][k] * ans[i + k]) % MOD;
  }
  printf("%lld", ans[k]);
  return 0;
}
