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;
int n, k;
long long jc[1003], inv[1003], f[1003][1003][3][3];
long long calc(int n, int m) {
  return jc[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
long long quickpow(long long num, int x) {
  long long ans = 1;
  long long base = num % 1000000007;
  while (x) {
    if (x & 1) ans = ans * base % 1000000007;
    x >>= 1;
    base = base * base % 1000000007;
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % 1000000007;
  for (int i = 0; i <= n; i++) inv[i] = quickpow(jc[i], 1000000007 - 2);
  f[1][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= n; j++)
      for (int a = 0; a <= 1; a++)
        for (int b = 0; b <= 1; b++)
          if (f[i][j][a][b]) {
            f[i + 1][j][b][0] += f[i][j][a][b];
            f[i + 1][j][b][0] %= 1000000007;
            f[i + 1][j + 1][b][1] += f[i][j][a][b];
            f[i + 1][j + 1][b][1] %= 1000000007;
            if (!a)
              f[i + 1][j + 1][b][0] += f[i][j][a][b],
                  f[i + 1][j + 1][b][0] %= 1000000007;
          }
  long long ans = 0;
  for (int i = k; i <= n; i++) {
    int t = i - k;
    long long sum = f[n + 1][i][1][0] + f[n + 1][i][0][0];
    sum %= 1000000007;
    if (t & 1)
      ans = (ans - calc(i, k) * jc[n - i] % 1000000007 * sum % 1000000007 +
             1000000007) %
            1000000007;
    else
      ans = (ans + calc(i, k) * jc[n - i] % 1000000007 * sum % 1000000007) %
            1000000007;
  }
  printf("%I64d\n", (ans % 1000000007 + 1000000007) % 1000000007);
}
