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;
const long long maxn = 1e3 + 10;
const long long mod = 1e9 + 7;
long long n, m;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
long long f[maxn][maxn][5];
long long F[maxn], ans;
long long jc[maxn], c[maxn][maxn];
signed main() {
  n = read(), m = read();
  jc[0] = f[1][0][0] = f[1][1][2] = 1;
  for (long long i = 2; i <= n; i++) {
    f[i][0][0] = 1;
    for (long long j = 1; j <= i; j++) {
      f[i][j][0] = (f[i - 1][j][0] + f[i - 1][j][1] + f[i - 1][j - 1][0]) % mod;
      f[i][j][1] = (f[i - 1][j][2] + f[i - 1][j][3] + f[i - 1][j - 1][2]) % mod;
      f[i][j][2] = (f[i - 1][j - 1][0] + f[i - 1][j - 1][1]) % mod;
      f[i][j][3] = (f[i - 1][j - 1][2] + f[i - 1][j - 1][3]) % mod;
    }
  }
  for (long long i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % mod;
  for (long long i = 0; i <= n; i++)
    F[i] = (f[n][i][0] + f[n][i][1]) * jc[n - i] % mod;
  c[0][0] = 1;
  for (long long i = 1; i <= n; i++) {
    c[i][0] = 1;
    for (long long j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  for (long long i = m; i <= n; i++)
    ans = (ans + ((i - m) & 1 ? -1 : 1) * c[i][m] * F[i] % mod) % mod;
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}
