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;
long long get() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const long long N = 1005, S = 1000, P = 1e9 + 7, sgn[] = {1, P - 1};
long long n, m, ans;
long long fac[S + 5], inv[S + 5];
long long f[N][N][2][2], g[N];
long long qpow(long long x, long long y) {
  long long res(1);
  while (y) res = res * ((y & 1) ? x : 1) % P, x = x * x % P, y >>= 1;
  return res;
}
void init() {
  fac[0] = 1;
  for (long long i = 1; i <= S; i++) fac[i] = fac[i - 1] * i % P;
  inv[S] = qpow(fac[S], P - 2);
  for (long long i = S; i >= 1; i--) inv[i - 1] = inv[i] * i % P;
}
long long C(long long x, long long y) {
  return x < y ? 0 : fac[x] * inv[y] % P * inv[x - y] % P;
}
signed main() {
  n = get(), m = get();
  init();
  f[1][0][0][0] = f[1][1][0][1] = 1;
  for (long long i = 2; i <= n; i++) {
    f[i][0][0][0] = 1;
    for (long long j = 1; j <= i; j++) {
      if (i != n) {
        (f[i][j][0][0] += f[i - 1][j - 1][0][0]) %= P;
        (f[i][j][1][0] += f[i - 1][j - 1][0][1]) %= P;
        (f[i][j][0][1] += f[i - 1][j - 1][0][0] + f[i - 1][j - 1][1][0]) %= P;
        (f[i][j][1][1] += f[i - 1][j - 1][0][1] + f[i - 1][j - 1][1][1]) %= P;
        (f[i][j][0][0] += f[i - 1][j][0][0] + f[i - 1][j][1][0]) %= P;
        (f[i][j][1][0] += f[i - 1][j][0][1] + f[i - 1][j][1][1]) %= P;
      } else {
        (f[i][j][0][0] += f[i - 1][j - 1][0][0]) %= P;
        (f[i][j][1][0] += f[i - 1][j - 1][0][1]) %= P;
        (f[i][j][0][0] += f[i - 1][j][0][0] + f[i - 1][j][1][0]) %= P;
        (f[i][j][1][0] += f[i - 1][j][0][1] + f[i - 1][j][1][1]) %= P;
      }
    }
  }
  for (long long i = 0; i <= n; i++)
    g[i] = (f[n][i][0][0] + f[n][i][1][0]) * fac[n - i] % P;
  for (long long i = m; i <= n; i++)
    ans = (ans + sgn[(i - m) & 1] * C(i, m) % P * g[i] % P) % P;
  printf("%lld\n", ans);
  return 0;
}
