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;
inline int Get_Int() {
  int Num = 0, Flag = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') Flag = -Flag;
  } while (ch < '0' || ch > '9');
  do {
    Num = Num * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return Num * Flag;
}
int N, M;
long long C[1005][1005], Factorial[1005], DP[1005][1005][4], Ans[1005];
inline void Prepare() {
  for (int i = 0; i <= N; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007;
  }
  Factorial[0] = 1;
  for (int i = 1; i <= N; ++i) Factorial[i] = Factorial[i - 1] * i % 1000000007;
}
int main() {
  cin >> N >> M;
  Prepare();
  DP[0][0][2] = 1;
  for (int i = 1; i <= N; ++i)
    for (int j = 0; j < i; ++j)
      for (int k = 0; k < 4; ++k) {
        long long Now = DP[i - 1][j][k];
        if (k & 1)
          DP[i][j + 1][k / 2 + 2] =
              (DP[i][j + 1][k / 2 + 2] + Now) % 1000000007;
        if (i < N)
          DP[i][j + 1][k / 2] = (DP[i][j + 1][k / 2] + Now) % 1000000007;
        DP[i][j][k / 2 + 2] = (DP[i][j][k / 2 + 2] + Now) % 1000000007;
      }
  for (int i = 0; i <= N; ++i) {
    for (int j = 0; j < 4; ++j) Ans[i] = (Ans[i] + DP[N][i][j]) % 1000000007;
    Ans[i] = Ans[i] * Factorial[N - i] % 1000000007;
  }
  for (int i = N; i >= 0; --i)
    for (int j = i + 1; j <= N; ++j)
      Ans[i] = (Ans[i] + -Ans[j] * C[j][i]) % 1000000007;
  Ans[M] = (Ans[M] + 1000000007) % 1000000007;
  cout << Ans[M] << endl;
  return 0;
}
