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;
inline long long read() {
  long long f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
const long long MAXN = 2e3 + 11;
long long N, M, f[MAXN][MAXN][2], fac[MAXN], C[MAXN][MAXN];
signed main() {
  N = read(), M = read();
  fac[0] = 1;
  for (long long i = 1; i <= N; i++) fac[i] = fac[i - 1] * i % 1000000007;
  C[0][0] = 1;
  for (long long i = 1; i <= N; i++) {
    C[i][0] = 1;
    for (long long j = 1; j <= N; j++)
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1], C[i][j] %= 1000000007;
  }
  f[0][0][0] = 1;
  for (long long i = 1; i <= 2 * N - 2; i++) {
    for (long long j = 0; j <= N; j++) {
      f[i][j][0] += f[i - 1][j][0] + f[i - 1][j][1];
      if (j && i != N) f[i][j][1] += f[i - 1][j - 1][0];
      if (j && i == N) f[i][j][1] += f[i - 1][j - 1][0] + f[i - 1][j - 1][1];
      f[i][j][0] %= 1000000007, f[i][j][1] %= 1000000007;
    }
  }
  long long Ans = 0, pw = 1;
  for (long long i = M; i <= N; i++) {
    long long W = ((f[2 * N - 2][i][0] + f[2 * N - 2][i][1]) * fac[N - i]) %
                  1000000007 * C[i][M] % 1000000007;
    Ans += pw * W, Ans %= 1000000007;
    pw *= -1;
    Ans = ((Ans % 1000000007) + 1000000007) % 1000000007;
  }
  printf("%lld\n", Ans);
}
