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 dp[1002][1002][2][2], jie[1002], ni[1002], g[1002], ans;
const int mod = 1e9 + 7;
long long calc(int n, int m) { return jie[n] * ni[m] % mod * ni[n - m] % mod; }
long long power(long long x, int y) {
  long long ans = 1;
  while (y) {
    if (y & 1) (ans *= x) %= mod;
    (x *= x) %= mod;
    y >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  jie[0] = 1;
  for (int i = 1; i <= n; ++i) jie[i] = (jie[i - 1] * i) % mod;
  ni[n] = power(jie[n], mod - 2);
  for (int i = n - 1; i >= 0; --i) ni[i] = ni[i + 1] * (i + 1) % mod;
  dp[0][0][1][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= n; ++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 - 1][j - 1][0][0]) %= mod;
        dp[i][j][0][1] +=
            (dp[i - 1][j - 1][0][0] + dp[i - 1][j - 1][1][0]) % mod;
        dp[i][j][0][1] %= mod;
        (dp[i][j][1][0] += dp[i - 1][j - 1][0][1]) %= mod;
        dp[i][j][1][1] +=
            (dp[i - 1][j - 1][0][1] + dp[i - 1][j - 1][1][1]) % mod;
        dp[i][j][1][1] %= mod;
      }
    }
  }
  for (int i = k; i <= n; ++i)
    g[i] = (dp[n][i][0][0] + dp[n][i][1][0]) % mod * jie[n - i] % mod;
  for (int i = k; i <= n; ++i)
    (ans += (((i - k) & 1) ? -1 : 1) * calc(i, k) * g[i] % mod + mod) %= mod;
  ans = (ans + mod) % mod;
  cout << ans;
  return 0;
}
