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;
template <class T>
bool ckmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <class T>
bool ckmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
const int N = 1005;
const int mod = 1e9 + 7;
int n, m, dp[N][N][2][2][2], C[N][N], ans, fac[N];
inline void fmod(int &x) { (x >= mod) && (x -= mod); }
signed main() {
  n = read(), m = read();
  dp[0][0][0][0][0] = 1;
  for (int i = 1, iend = n; i <= iend; ++i) {
    for (int j = 0, jend = i; j <= jend; ++j) {
      fmod(dp[i][j][0][0][0] = dp[i - 1][j][0][0][0] + dp[i - 1][j][1][0][0]);
      if (j > 0)
        fmod(dp[i][j][0][0][1] =
                 dp[i - 1][j - 1][0][0][0] + dp[i - 1][j - 1][1][0][0]);
      fmod(dp[i][j][0][1][0] = dp[i - 1][j][0][0][1] + dp[i - 1][j][1][0][1]);
      if (j > 0)
        fmod(dp[i][j][0][1][1] =
                 dp[i - 1][j - 1][0][0][1] + dp[i - 1][j - 1][1][0][1]);
      fmod(dp[i][j][1][0][0] = dp[i - 1][j][0][1][0] + dp[i - 1][j][1][1][0]);
      if (i > 1 && j > 0)
        fmod(dp[i][j][1][0][0] += dp[i - 1][j - 1][0][0][0]),
            fmod(dp[i][j][1][0][0] += dp[i - 1][j - 1][1][0][0]);
      if (i > 1 && j > 0)
        fmod(dp[i][j][1][0][1] =
                 dp[i - 1][j - 1][0][1][0] + dp[i - 1][j - 1][1][1][0]);
      fmod(dp[i][j][1][1][0] = dp[i - 1][j][0][1][1] + dp[i - 1][j][1][1][1]);
      if (i > 1 && j > 0)
        fmod(dp[i][j][1][1][0] += dp[i - 1][j - 1][0][0][1]),
            fmod(dp[i][j][1][1][0] += dp[i - 1][j - 1][1][0][1]);
      if (i > 1 && j > 0)
        fmod(dp[i][j][1][1][1] =
                 dp[i - 1][j - 1][0][1][1] + dp[i - 1][j - 1][1][1][1]);
    }
  }
  for (int i = 0, iend = n; i <= iend; ++i) {
    C[i][0] = 1;
    for (int j = 1, jend = i; j <= jend; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
  }
  fac[0] = 1;
  for (int i = 1, iend = n + 1; i <= iend; ++i)
    fac[i] = 1ll * i * fac[i - 1] % mod;
  for (int i = m, iend = n; i <= iend; ++i) {
    int s = 0;
    for (int a = 0, aend = 1; a <= aend; ++a)
      for (int b = 0, bend = 1; b <= bend; ++b)
        s = (s + dp[n][i][a][b][0]) % mod;
    int tmp = (long long)C[i][m] * s % mod * fac[n - i] % mod;
    if ((i - m) & 1) tmp = mod - tmp;
    ans = (ans + tmp) % mod;
  }
  cout << ans << '\n';
}
