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;
int n, m;
long long fac[1005] = {1}, inv[1005], f[1005][1005][2][2], g[1005], ans;
const long long mod = 1e9 + 7;
inline int read() {
  int w = 0, f = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    w = (w << 1) + (w << 3) + ch - '0', ch = getchar();
  return w * f;
}
long long ksm(long long a, long long p) {
  long long ret = 1ll;
  while (p) {
    if (p & 1) (ret *= a) %= mod;
    p >>= 1, (a *= a) %= mod;
  }
  return ret;
}
long long C(int x, int y) { return fac[x] * inv[y] % mod * inv[x - y] % mod; }
int main() {
  for (int i = 1; i <= 1000; i++) fac[i] = fac[i - 1] * i % mod;
  inv[1000] = ksm(fac[1000], mod - 2);
  for (int i = 999; ~i; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  n = read(), m = read(), f[1][0][0][0] = f[1][1][0][1] = f[n][0][0][0] = 1;
  for (int i = 2; i <= n; i++) {
    f[i][0][0][0] = 1;
    for (int j = 1; j <= i; j++) {
      f[i][j][0][0] =
          (f[i - 1][j][1][0] + f[i - 1][j][0][0] + f[i - 1][j - 1][0][0]) % mod;
      f[i][j][0][1] = (f[i - 1][j - 1][0][0] + f[i - 1][j - 1][1][0]) % mod;
      f[i][j][1][0] =
          (f[i - 1][j - 1][0][1] + f[i - 1][j][0][1] + f[i - 1][j][1][1]) % mod;
      f[i][j][1][1] = (f[i - 1][j - 1][0][1] + f[i - 1][j - 1][1][1]) % mod;
    }
  }
  for (int i = 0; i <= n; i++)
    g[i] = (f[n][i][0][0] + f[n][i][1][0]) * fac[n - i] % mod;
  for (int i = m; i <= n; i++)
    if ((i - m) & 1)
      (ans += mod - C(i, m) * g[i] % mod) %= mod;
    else
      (ans += C(i, m) * g[i] % mod) %= mod;
  printf("%lld\n", ans);
  return 0;
}
