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;
using ll = long long;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
const int mod = 1000000007;
int dp[maxn][maxn][2][2];
ll f[maxn];
ll jc[maxn];
ll qp(ll a, ll b) {
  ll ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ans;
}
ll C(int x, int y) {
  return jc[x] * qp(jc[y], mod - 2) % mod * qp(jc[x - y], mod - 2) % mod;
}
int n, k;
int main() {
  cin >> n >> k;
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % mod;
  dp[0][0][1][0] = 1;
  for (int i = 0; i <= n - 1; i++) {
    for (int j = 0; j <= n; j++) {
      for (int a = 0; a <= 1; a++) {
        for (int b = 0; b <= 1; b++) {
          int x = dp[i][j][a][b];
          if (x == 0) continue;
          (dp[i + 1][j][b][0] += x) %= mod;
          if (a == 0) (dp[i + 1][j + 1][b][0] += x) %= mod;
          (dp[i + 1][j + 1][b][1] += x) %= mod;
        }
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    f[i] = jc[n - i] * (dp[n][i][0][0] + dp[n][i][1][0]) % mod;
  }
  ll ans = 0;
  int flag = 1;
  for (int i = k; i <= n; i++) {
    ans += flag * f[i] * C(i, k) % mod;
    flag *= -1;
  }
  cout << (ans % mod + mod) % mod << "\n";
  return 0;
}
