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>
const long long MOD = 1e9 + 7;
long long fac[2050], inv[2050];
long long f[2050][2050][2][2];
long long QPow(long long x, long long y) {
  long long ans = 1, base = x % MOD;
  while (y) {
    if (y & 1) {
      ans = (ans * base) % MOD;
    }
    base = (base * base) % MOD;
    y >>= 1;
  }
  return ans;
}
long long Combine(int m, int n) {
  return m > n ? 0 : (((fac[n] * inv[n - m]) % MOD) * inv[m]) % MOD;
}
int main() {
  int n, m;
  long long ans = 0;
  scanf("%d%d", &n, &m);
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = (fac[i - 1] * i) % MOD;
  }
  inv[n] = QPow(fac[n], MOD - 2);
  for (int i = n - 1; i >= 1; i--) {
    inv[i] = (inv[i + 1] * (i + 1)) % MOD;
  }
  f[1][0][0][0] = f[1][1][0][1] = 1;
  for (int i = 2; i <= n; i++) {
    f[i][0][0][0] = 1;
    for (int j = 1; j <= i; j++) {
      f[i][j][0][0] = (((f[i - 1][j][1][0] + f[i - 1][j][0][0]) % MOD) +
                       f[i - 1][j - 1][0][0]) %
                      MOD;
      f[i][j][0][1] = (f[i - 1][j - 1][0][0] + f[i - 1][j - 1][1][0]) % MOD;
      f[i][j][1][0] = (((f[i - 1][j - 1][0][1] + f[i - 1][j][1][1]) % MOD) +
                       f[i - 1][j][0][1]) %
                      MOD;
      f[i][j][1][1] = (f[i - 1][j - 1][0][1] + f[i - 1][j - 1][1][1]) % MOD;
    }
  }
  f[n][0][0][0] = 1;
  for (int i = 1; i <= n; i++) {
    f[n][i][0][0] = (((f[n - 1][i][1][0] + f[n - 1][i][0][0]) % MOD) +
                     f[n - 1][i - 1][0][0]) %
                    MOD;
    f[n][i][1][0] = ((f[n - 1][i - 1][0][1] + f[n - 1][i][1][1] % MOD) +
                     f[n - 1][i][0][1]) %
                    MOD;
  }
  for (int i = m; i <= n; i++) {
    long long tmp =
        (Combine(m, i) *
         (fac[n - i] * ((f[n][i][0][0] + f[n][i][1][0]) % MOD) % MOD) % MOD) %
        MOD;
    ans = (ans + ((i - m) & 1 ? MOD - tmp : tmp)) % MOD;
  }
  printf("%lld", ans);
  return 0;
}
