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 int INF = ((long long)1E9 + 7);
template <class C>
void mini(C& a, C b) {
  a = min(a, b);
}
template <class C>
void maxi(C& a, C b) {
  a = max(a, b);
}
template <class C>
void add(C& a, C b) {
  a = (a + b) % INF;
}
const int MN = 1002;
int T[MN][MN][4];
int bin[MN][MN];
int A[MN];
long long B[MN];
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  bin[0][0] = 1;
  for (int i = (1); i <= (n); ++i) {
    bin[i][0] = 1;
    for (int j = (1); j <= (n); ++j)
      bin[i][j] = (bin[i - 1][j - 1] + bin[i - 1][j]) % INF;
  }
  T[0][0][1] = 1;
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (n); ++j) {
      for (int m = 0; m < (4); ++m) {
        T[i][j][m] %= INF;
        int v = T[i][j][m];
        if ((m & 1) == 0) add(T[i + 1][j + 1][m >> 1], v);
        add(T[i + 1][j][m >> 1], v);
        add(T[i + 1][j + 1][(m >> 1) | 2], v);
      }
    }
  for (int i = (0); i <= (n); ++i) {
    A[i] = (T[n][i][0] + T[n][i][1]) % INF;
  }
  B[n] = A[n];
  long long f = 1;
  for (int i = (n - 1); i >= (0); --i) {
    f = (f * (n - i)) % INF;
    B[i] = (f * A[i]) % INF;
    for (int j = (i + 1); j <= (n); ++j) {
      B[i] = (B[i] - B[j] * bin[j][i]) % INF;
    }
    B[i] = (B[i] + INF) % INF;
  }
  cout << B[k] << endl;
  return 0;
}
