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;
void add(int &a, int b) { a += b, a %= 1000000007; }
int dp[1000 + 5][1000 + 5][2][2], n, g, c[1000 + 5][1000 + 5], ans[1000 + 5];
int main() {
  for (int i = 0; i <= 1000; i++)
    for (int j = 0; j <= i; j++) {
      if (!i) c[i][j] = 1;
      if (i) add(c[i][j], c[i - 1][j]);
      if (j) add(c[i][j], c[i - 1][j - 1]);
    }
  scanf("%d%d", &n, &g);
  dp[0][0][0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++)
      for (int k = 0; k < 2; k++)
        for (int l = 0; l < 2; l++)
          if (dp[i - 1][j][k][l]) {
            if (!k && i - 1) add(dp[i][j + 1][l][0], dp[i - 1][j][k][l]);
            if (i < n) add(dp[i][j + 1][l][1], dp[i - 1][j][k][l]);
            add(dp[i][j][l][0], dp[i - 1][j][k][l]);
          }
  for (int i = 0; i <= n; i++) {
    for (int k = 0; k < 2; k++)
      for (int l = 0; l < 2; l++) add(ans[i], dp[n][i][k][l]);
    for (int j = 2; j <= n - i; j++) ans[i] = ans[i] * 1ll * j % 1000000007;
  }
  for (int i = n - 1; i >= g; i--)
    for (int j = i + 1; j <= n; j++)
      ans[i] -= ans[j] * 1ll * c[j][i] % 1000000007, ans[i] %= 1000000007;
  if (ans[g] < 0) ans[g] += 1000000007;
  printf("%d\n", ans[g]);
  return 0;
}
