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 int N = 1010, mod = 1000000007;
int n, m;
long long ans, f[N][N][4], jc[N], inv[N], g[N];
long long power(long long a, long long b = mod - 2) {
  long long c = 1;
  while (b) {
    if (b & 1) c = c * a % mod;
    b /= 2;
    a = a * a % mod;
  }
  return c;
}
long long C(int x, int y) { return jc[x] * inv[y] % mod * inv[x - y] % mod; }
int main() {
  scanf("%d %d", &n, &m);
  f[1][1][1] = f[1][0][0] = 1;
  for (int i = 2; i <= n; i++) {
    f[i][0][0] = 1;
    for (int j = 1; j <= i; j++)
      f[i][j][0] = (f[i - 1][j - 1][0] + f[i - 1][j][0] + f[i - 1][j][2]) % mod,
      f[i][j][2] = (f[i - 1][j - 1][1] + f[i - 1][j][1] + f[i - 1][j][3]) % mod,
      f[i][j][1] = (f[i - 1][j - 1][0] + f[i - 1][j - 1][2]) % mod,
      f[i][j][3] = (f[i - 1][j - 1][1] + f[i - 1][j - 1][3]) % mod;
  }
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = jc[i - 1] * i % mod;
  inv[n] = power(jc[n]);
  for (int i = n; i; i--) inv[i - 1] = inv[i] * i % mod;
  for (int i = m, F = 1; i <= n; i++) {
    g[i] = (f[n][i][2] + f[n][i][0]) * jc[n - i] % mod;
    (ans += F * C(i, m) * g[i]) %= mod;
    F = -F;
  }
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}
