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 mod = 1000000007;
int dp[1010][1010][2][2];
long long f[1010], mi[1010], ni[1010];
long long powmod(long long a, int b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a, ans %= mod;
    a *= a, b >>= 1;
    a %= mod;
  }
  return ans;
}
long long C(int i, int j) { return mi[i] * ni[j] % mod * ni[i - j] % mod; }
void upd(int &n, int k) {
  n += k;
  if (n >= mod) n -= mod;
}
int main() {
  int i, j, is1, is2, k, t, p, n;
  dp[0][0][1][0] = 1;
  for (i = 1, mi[0] = 1; i <= 1000; i++) mi[i] = mi[i - 1] * i, mi[i] %= mod;
  for (i = 0; i <= 1000; i++) ni[i] = powmod(mi[i], mod - 2);
  scanf("%d%d", &n, &p);
  for (i = 0; i < n; i++) {
    for (j = 0; j <= i; j++)
      for (is1 = 0; is1 < 2; is1++)
        for (is2 = 0; is2 < 2; is2++)
          if (dp[i][j][is1][is2]) {
            k = dp[i][j][is1][is2];
            upd(dp[i + 1][j][is2][0], k);
            if (!is1) upd(dp[i + 1][j + 1][is2][0], k);
            if (i < n - 1) upd(dp[i + 1][j + 1][is2][1], k);
          };
  }
  for (i = 0; i <= n; i++) f[i] = dp[n][i][0][0] + dp[n][i][1][0], f[i] %= mod;
  for (i = 0; i <= n; i++) f[i] *= mi[n - i], f[i] %= mod;
  for (i = n - 1; i >= p; i--) {
    for (j = i + 1; j <= n; j++) {
      f[i] -= f[j] * C(j, i);
      f[i] %= mod;
    }
  }
  f[p] += mod;
  f[p] %= mod;
  printf("%I64d\n", f[p]);
}
