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 mod = 1e9 + 7;
long long dp[1010][1010][2][2], jie[1010], res, c[1010][1010], ans[1010];
int main() {
  scanf("%d%d", &n, &k);
  if (n <= 2) {
    if (n == 1) {
      if (k == 0)
        printf("1\n");
      else
        printf("0\n");
    } else {
      if (k == 0 || k == 2)
        printf("1\n");
      else
        printf("0\n");
    }
    return 0;
  }
  c[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= i; ++j)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
  }
  jie[0] = 1;
  for (int i = 1; i <= n; ++i) jie[i] = jie[i - 1] * i % mod;
  dp[1][0][0][0] = 1;
  dp[1][1][0][1] = 1;
  for (int i = 2; i <= n; ++i) {
    for (int j = 0; j <= i; ++j) {
      dp[i][j][0][0] = (dp[i - 1][j][0][0] + dp[i - 1][j][1][0]) % mod;
      dp[i][j][1][0] = (dp[i - 1][j][0][1] + dp[i - 1][j][1][1]) % mod;
      if (j) {
        dp[i][j][0][0] = (dp[i][j][0][0] + dp[i - 1][j - 1][0][0]) % mod;
        dp[i][j][1][0] = (dp[i][j][1][0] + dp[i - 1][j - 1][0][1]) % mod;
        dp[i][j][0][1] =
            (dp[i][j][0][1] + dp[i - 1][j - 1][0][0] + dp[i - 1][j - 1][1][0]) %
            mod;
        dp[i][j][1][1] =
            (dp[i][j][1][1] + dp[i - 1][j - 1][0][1] + dp[i - 1][j - 1][1][1]) %
            mod;
      }
    }
  }
  for (int j = k; j <= n; ++j)
    ans[j] = (dp[n][j][0][0] + dp[n][j][1][0]) % mod * jie[n - j] % mod;
  for (int i = n; i >= k; --i) {
    for (int j = n; j > i; --j)
      ans[i] = (ans[i] - c[j][i] * ans[j] % mod + mod) % mod;
  }
  printf("%lld\n", ans[k]);
  return 0;
}
