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 = 2005;
const long long mo = 1e9 + 7;
long long dp[maxn][maxn][2][2], n, k;
long long ce[maxn][maxn], ans[maxn], f[maxn];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    ce[i][0] = ce[i][i] = 1LL;
    for (int j = 1; j <= i - 1; j++) {
      ce[i][j] = (ce[i][j] + ce[i - 1][j] + ce[i - 1][j - 1]) % mo;
    }
  }
  f[0] = f[1] = 1LL;
  for (int i = 2; i <= 2001; i++) f[i] = (f[i - 1] * i * 1LL) % mo;
  dp[0][0][1][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= i - 1; j++)
      for (int n1 = 0; n1 <= 1; n1++)
        for (int n2 = 0; n2 <= 1; n2++) {
          dp[i][j][n2][0] = (dp[i][j][n2][0] + dp[i - 1][j][n1][n2]) % mo;
          if (!n1)
            dp[i][j + 1][n2][0] =
                (dp[i][j + 1][n2][0] + dp[i - 1][j][n1][n2]) % mo;
          if (i < n)
            dp[i][j + 1][n2][1] =
                (dp[i][j + 1][n2][1] + dp[i - 1][j][n1][n2]) % mo;
        }
  }
  for (int i = 0; i <= n; i++) {
    for (int n1 = 0; n1 <= 1; n1++)
      for (int n2 = 0; n2 <= 1; n2++) ans[i] = (ans[i] + dp[n][i][n1][n2]) % mo;
    ans[i] = ans[i] * f[n - i] % mo;
  }
  int o = -1;
  for (int i = k + 1; i <= n; i++) {
    ans[k] = (ans[k] + o * ce[i][k] * ans[i] + mo) % mo;
    o *= -1;
  }
  cout << ans[k];
}
