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 gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - 48;
    ch = getchar();
  }
  return x * f;
}
template <typename T>
inline bool Max(T& a, const T& b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool Min(T& a, const T& b) {
  return a > b ? a = b, true : false;
}
const int N = 1e3 + 7, mod = 1e9 + 7;
template <typename T>
inline void add(T& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int n, m;
int f[N][N][2][2], inv[N], fac[N], invfac[N];
int C(int n, int m) {
  if (n < m) return 0;
  return 1ll * fac[n] * invfac[n - m] % mod * invfac[m] % mod;
}
int main() {
  n = gi(), m = gi();
  f[1][0][1][0] = 1;
  invfac[0] = fac[0] = inv[1] = 1;
  for (int i = 1; i <= n; i++) {
    if (i ^ 1) inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    fac[i] = 1ll * fac[i - 1] * i % mod;
    invfac[i] = 1ll * invfac[i - 1] * inv[i] % mod;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i; j++)
      for (int a = 0; a <= 1; a++)
        for (int b = 0; b <= 1; b++)
          if (f[i][j][a][b]) {
            add(f[i + 1][j][b][0], f[i][j][a][b]);
            add(f[i + 1][j + 1][b][1], f[i][j][a][b]);
            if (!a) add(f[i + 1][j + 1][b][0], f[i][j][a][b]);
          }
  int ans = 0;
  for (int i = m; i <= n; i++) {
    int c = ((i - m) & 1 ? -1 : 1);
    add(ans,
        (1ll * c * C(i, m) % mod * (f[n + 1][i][0][0] + f[n + 1][i][1][0]) %
             mod * fac[n - i] % mod +
         mod) %
            mod);
  }
  printf("%d\n", ans);
  return 0;
}
