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>
const int mod = 1000000007;
int n, k;
long long dp[1001][1001][2][2], f[1001], c[1001][1001], r[1001][1001];
int getans(int k) {
  long long ans = 0;
  for (int i = k; i <= n; i++)
    ans = (ans + c[i][k] * f[i] * ((i - k) & 1 ? -1 : 1) + mod) % mod;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  if (n == 1) {
    printf("%d\n", k == 0);
    return 0;
  }
  for (int i = 0; i <= n; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  dp[1][1][0][1] = r[1][1] = 1;
  for (int i = 2; i <= n; i++) {
    dp[i][1][0][0] = 1;
    if (i < n) dp[i][1][0][1] = 1;
    for (int j = 2; j <= i; j++) {
      dp[i][j][0][0] = (r[i - 1][j - 1] -
                        (dp[i - 1][j - 1][0][1] + dp[i - 1][j - 1][1][0] +
                         dp[i - 1][j - 1][1][1] + dp[i - 2][j - 1][0][1] +
                         dp[i - 2][j - 1][1][1]) %
                            mod +
                        mod) %
                       mod;
      dp[i][j][1][0] = dp[i - 1][j - 1][0][1];
      if (i < n) {
        dp[i][j][0][1] = (r[i - 1][j - 1] - dp[i - 1][j - 1][0][1] -
                          dp[i - 1][j - 1][1][1] + mod * 2) %
                         mod;
        dp[i][j][1][1] =
            (dp[i - 1][j - 1][0][1] + dp[i - 1][j - 1][1][1]) % mod;
      }
    }
    for (int j = 1; j <= i; j++)
      r[i][j] = (r[i - 1][j] + dp[i][j][0][0] + dp[i][j][0][1] +
                 dp[i][j][1][0] + dp[i][j][1][1]) %
                mod;
  }
  for (int j = 1; j <= n; j++) f[j] = r[n][j];
  long long tmp = 1;
  for (int j = 1; j <= n; j++) {
    tmp = tmp * j % mod;
    f[n - j] = f[n - j] * tmp % mod;
  }
  if (k == 0) {
    for (int i = 1; i <= n; i++) tmp = (tmp - getans(i) + mod) % mod;
    printf("%d\n", (int)tmp);
  } else
    printf("%d\n", getans(k));
}
