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 mod = 1e9 + 7, maxn = 2e3 + 5;
long long f[maxn][maxn][2][2];
long long g[maxn];
long long n, m;
long long upd(const long long &x) {
  if (x < 0) return x + mod;
  if (x >= mod) return x - mod;
  return x;
}
void add(long long &x, const long long &y) { x = upd(x + upd(y)); }
long long jc[maxn], inv[maxn];
long long ksm(long long x, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * x % mod;
    x = x * x % mod;
    k >>= 1;
  }
  return ans;
}
void init(long long n) {
  jc[0] = 1;
  for (long long i(1), endi(n); i <= endi; i++) jc[i] = jc[i - 1] * i % mod;
  inv[n] = ksm(jc[n], mod - 2);
  for (long long i(n - 1), endi(0); i >= endi; i--)
    inv[i] = inv[i + 1] * (i + 1) % mod;
}
long long C(long long n, long long m) {
  if (n < 0 || m < 0 || n < m) return 0;
  return jc[n] * inv[m] % mod * inv[n - m] % mod;
}
signed main() {
  scanf("%lld%lld", &n, &m);
  init(n);
  f[1][0][0][0] = 1;
  f[1][1][0][1] = 1;
  for (long long i(2), endi(n); i <= endi; i++)
    for (long long j(0), endj(n); j <= endj; j++) {
      if (j) {
        add(f[i][j][0][0], f[i - 1][j - 1][0][0]);
        add(f[i][j][1][0], f[i - 1][j - 1][0][1]);
        if (i != n) {
          add(f[i][j][0][1], f[i - 1][j - 1][0][0] + f[i - 1][j - 1][1][0]);
          add(f[i][j][1][1], f[i - 1][j - 1][0][1] + f[i - 1][j - 1][1][1]);
        }
      }
      add(f[i][j][0][0], f[i - 1][j][0][0] + f[i - 1][j][1][0]);
      add(f[i][j][1][0], f[i - 1][j][0][1] + f[i - 1][j][1][1]);
    }
  for (long long i(0), endi(n); i <= endi; i++) {
    add(g[i], f[n][i][0][0] + f[n][i][1][0]);
    g[i] = g[i] * jc[n - i] % mod;
  }
  long long ans = 0;
  for (long long i(m), endi(n); i <= endi; i++) {
    add(ans, ((i - m) & 1 ? -1 : 1) * g[i] * C(i, m) % mod);
  }
  cout << ans;
  return 0;
}
