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 N = 1000 + 7, Mod = 1000000000 + 7;
int n, m, Dp[N][N][2][2], Ans[N], C[N][N], Fac[N];
inline int Inc(int A, int B) {
  A += B;
  return A >= Mod ? A - Mod : A;
}
inline int Mul(int A, int B) { return 1ll * A * B % Mod; }
inline int Dec(int A, int B) {
  A -= B;
  return A < 0 ? A + Mod : A;
}
inline void Init(int n) {
  for (int i = (0), iLimit = (n); i <= iLimit; (++(i)))
    for (int j = (0), jLimit = (i); j <= jLimit; (++(j)))
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = Inc(C[i - 1][j], C[i - 1][j - 1]);
  Fac[0] = 1;
  for (int i = (1), iLimit = (n); i <= iLimit; (++(i)))
    Fac[i] = Mul(Fac[i - 1], i);
}
int main() {
  cin >> n >> m;
  Init(N - 1);
  Dp[0][0][1][0] = 1;
  for (int i = (1), iLimit = (n); i <= iLimit; (++(i)))
    for (int j = (0), jLimit = (i); j <= jLimit; (++(j)))
      for (int l = (0), lLimit = (1); l <= lLimit; (++(l)))
        for (int r = (0), rLimit = (1); r <= rLimit; (++(r))) {
          Dp[i][j][l][0] = Inc(Dp[i][j][l][0], Dp[i - 1][j][r][l]);
          if (!r && j)
            Dp[i][j][l][0] = Inc(Dp[i][j][l][0], Dp[i - 1][j - 1][r][l]);
          if (j) Dp[i][j][l][1] = Inc(Dp[i][j][l][1], Dp[i - 1][j - 1][r][l]);
        }
  for (int i = (0), iLimit = (n); i <= iLimit; (++(i)))
    Ans[i] = Mul(Inc(Dp[n][i][0][0], Dp[n][i][1][0]), Fac[n - i]);
  int Ret = Ans[m];
  for (int i = (m + 1), iLimit = (n); i <= iLimit; (++(i))) {
    if ((i - m) & 1)
      Ret = Dec(Ret, Mul(Ans[i], C[i][m]));
    else
      Ret = Inc(Ret, Mul(Ans[i], C[i][m]));
  }
  cout << Ret;
  return 0;
}
