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>
const int mod = 1e9 + 7;
int f[1010][1010][4], A[1010], a[1010], pw[1010], inp[1010];
inline int qpow(int base, int power) {
  int ret = 1;
  for (; power; power >>= 1, base = 1ll * base * base % mod)
    power& 1 ? ret = 1ll * ret * base % mod : 0;
  return ret;
}
inline int C(int n, int m) {
  return 1ll * pw[n] * inp[m] % mod * inp[n - m] % mod;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  f[0][0][2] = 1;
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) pw[i] = 1ll * pw[i - 1] * i % mod;
  inp[n] = qpow(pw[n], mod - 2);
  for (int i = n; i; --i) inp[i - 1] = 1ll * inp[i] * i % mod;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < i; ++j)
      for (int k = 0; k < 4; ++k) {
        if (!(k >> 1 & 1)) (f[i][j + 1][k << 1 & 3] += f[i - 1][j][k]) %= mod;
        (f[i][j][k << 1 & 3] += f[i - 1][j][k]) %= mod;
        if (i < n) (f[i][j + 1][k << 1 & 3 | 1] += f[i - 1][j][k]) %= mod;
      }
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j < 4; ++j)
      A[i] = (A[i] + 1ll * f[n][i][j] * pw[n - i]) % mod;
  int ans = 0;
  for (int i = k; i <= n; ++i)
    if ((i - k) & 1)
      (ans += mod - 1ll * C(i, k) * A[i] % mod) %= mod;
    else
      ans = (ans + 1ll * C(i, k) * A[i]) % mod;
  printf("%d\n", ans);
  return 0;
}
