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;
const long long inf = 1000000007;
long long a[1100][1100];
long long b[1100][1100];
long long A(int n, int k);
long long B(int n, int k);
long long A(int n, int k) {
  if (a[n][k] != -1) {
    return a[n][k];
  }
  if (n < k) {
    return a[n][k] = 0;
  }
  if (n == 1) {
    return a[n][k] = (k == 0);
  }
  if (n == 2) {
    return a[n][k] = (k != 1);
  }
  long long ret = A(n - 1, k);
  ret += A(n - 1, k + 1) * (k + 1) - B(n - 1, k + 1) + B(n - 1, k) -
         B(n - 1, k + 1) + B(n - 1, k);
  ret += A(n - 1, k) * (n - 1 - k);
  ret -= A(n - 1, k) - A(n - 2, k) - B(n - 1, k);
  if (k > 0) {
    ret += A(n - 1, k - 1) - A(n - 2, k - 1) - B(n - 1, k - 1);
  }
  ret -= A(n - 1, k) - A(n - 2, k) - B(n - 1, k);
  if (k > 0) {
    ret += A(n - 1, k - 1) - A(n - 2, k - 1) - B(n - 1, k - 1);
  }
  ret -= A(n - 2, k);
  if (k > 1) {
    ret += A(n - 2, k - 2);
  }
  return a[n][k] = (ret % inf + inf) % inf;
}
long long B(int n, int k) {
  if (b[n][k] != -1) {
    return b[n][k];
  }
  if (n < k) {
    return (b[n][k] = 0);
  }
  if (n == 1) {
    return b[n][k] = 0;
  }
  if (n == 2) {
    return b[n][k] = (k == 2);
  }
  if (k == 0) {
    return b[n][k] = 0;
  }
  if (k == 1) {
    return b[n][k] =
               ((A(n - 1, 0) - A(n - 2, 0) + B(n - 1, 1)) % inf + inf) % inf;
  }
  return b[n][k] = (A(n - 2, k - 2) + B(n - 1, k) + A(n - 1, k - 1) -
                    B(n - 1, k - 1) - A(n - 2, k - 1)) %
                   inf;
}
int main() {
  memset(a, -1, sizeof(a));
  memset(b, -1, sizeof(b));
  int n, k;
  cin >> n >> k;
  cout << A(n, k) << endl;
  return 0;
}
