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;
long long c[1001][1001], p[1005];
long long dp[1001][1001][2][2], tmp;
long long num[1005], tim[1005], ans;
int n, k;
int main() {
  cin >> n >> k;
  c[0][0] = p[0] = 1;
  for (int i = 1; i <= n; i++) {
    c[i][0] = 1;
    p[i] = p[i - 1] * i % 1000000007;
    for (int j = 1; j <= n; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  }
  dp[1][1][0][1] = 1;
  dp[1][0][0][0] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= i; j++) {
      if (tmp = dp[i][j][0][0]) {
        (dp[i + 1][j][0][0] += tmp) %= 1000000007;
        (dp[i + 1][j + 1][0][0] += tmp) %= 1000000007;
        (dp[i + 1][j + 1][0][1] += tmp) %= 1000000007;
      }
      if (tmp = dp[i][j][0][1]) {
        (dp[i + 1][j][1][0] += tmp) %= 1000000007;
        (dp[i + 1][j + 1][1][0] += tmp) %= 1000000007;
        (dp[i + 1][j + 1][1][1] += tmp) %= 1000000007;
      }
      if (tmp = dp[i][j][1][0]) {
        (dp[i + 1][j][0][0] += tmp) %= 1000000007;
        (dp[i + 1][j + 1][0][1] += tmp) %= 1000000007;
      }
      if (tmp = dp[i][j][1][1]) {
        (dp[i + 1][j][1][0] += tmp) %= 1000000007;
        (dp[i + 1][j + 1][1][1] += tmp) %= 1000000007;
      }
    }
  for (int i = k; i <= n; i++)
    num[i] = (dp[n][i][0][0] + dp[n][i][1][0]) * p[n - i] % 1000000007;
  tim[k] = 1;
  for (int i = k; i <= n; i++) {
    ans = (ans + tim[i] * num[i]) % 1000000007;
    for (int j = i + 1; j <= n; j++)
      tim[j] =
          (tim[j] - tim[i] * c[j][i] % 1000000007 + 1000000007) % 1000000007;
  }
  printf("%I64d\n", ans);
}
