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 long long md = 1e9 + 7;
long long n, m, c[1010][1010], dp[1010][1010][1 << 2], cc[1010];
int main() {
  cin >> n >> m;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0) {
        c[i][j] = 1;
      } else {
        c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
        c[i][j] %= md;
      }
    }
  }
  dp[0][0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      for (int k = 0; k < 4; k++) {
        if (i > 1 && j > 0 && (k & 1) == 0) {
          dp[i][j][k >> 1] += dp[i - 1][j - 1][k];
          dp[i][j][k >> 1] %= md;
        }
        if (i < n && j > 0) {
          dp[i][j][(k >> 1) | 2] += dp[i - 1][j - 1][k];
          dp[i][j][(k >> 1) | 2] %= md;
        }
        dp[i][j][k >> 1] += dp[i - 1][j][k];
        dp[i][j][k >> 1] %= md;
      }
    }
  }
  long long nw = 1;
  for (int i = n; i >= 0; i--) {
    cc[i] = (dp[n][i][0] + dp[n][i][1]) * nw % md;
    nw *= (n - i + 1);
    nw %= md;
  }
  for (int i = n; i >= m; i--) {
    for (int j = n; j > i; j--) {
      cc[i] -= cc[j] * c[j][i] % md;
      cc[i] += md;
      cc[i] %= md;
    }
  }
  cout << cc[m] << endl;
  return 0;
}
