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 read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long n, k, f[2005][2005][2], C[2005][2005], fac[2005];
signed main() {
  n = read();
  k = read();
  fac[0] = 1;
  for (long long i = 0; i <= n; i++) {
    C[i][0] = 1;
    if (i != 0) fac[i] = fac[i - 1] * i % (long long)(1e9 + 7);
    for (long long j = 1; j <= i; j++)
      C[i][j] = C[i - 1][j] + C[i - 1][j - 1], C[i][j] %= (long long)(1e9 + 7);
  }
  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] %= (long long)(1e9 + 7);
      f[i][j][1] %= (long long)(1e9 + 7);
    }
  }
  long long poww = 1, ans = 0;
  for (long long i = k; i <= n; i++) {
    ans += poww *
           ((f[2 * n - 2][i][0] + f[2 * n - 2][i][1]) * fac[n - i] %
            (long long)(1e9 + 7)) *
           C[i][k] % (long long)(1e9 + 7);
    poww *= -1;
    ans %= (long long)(1e9 + 7);
    ans = (ans + (long long)(1e9 + 7)) % (long long)(1e9 + 7);
  }
  printf("%lld\n", ans);
  return 0;
}
