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 int MAXN = 1000 + 10, MOD = (int)1e9 + 7;
int n, k, f[MAXN][MAXN][2], g[MAXN][MAXN][2], h[MAXN], Tohka[MAXN], fac[MAXN],
    c[MAXN][MAXN];
void Plus(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void Dp(int f[][MAXN][2], int a, int b) {
  f[1][0][0] = a, f[1][1][1] = b;
  for (int i = 2; i <= (n + 1) / 2; ++i)
    for (int j = 0; j <= i; ++j) {
      f[i][j][0] = f[i - 1][j][0];
      Plus(f[i][j][0], f[i - 1][j][1]);
      if (j) {
        Plus(f[i][j][0], f[i - 1][j - 1][0]);
        Plus(f[i][j][1], f[i - 1][j - 1][0]);
        Plus(f[i][j][1], f[i - 1][j - 1][1]);
      }
    }
}
int main() {
  cin >> n >> k;
  f[0][0][0] = g[0][0][0] = 1;
  Dp(f, 1, 1);
  if (n & 1) Dp(g, 1, 0);
  fac[0] = 1;
  c[0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      c[i][j] = j ? (c[i - 1][j] + c[i - 1][j - 1]) % MOD : 1;
  for (int i = 1; i <= n; ++i) fac[i] = (long long)fac[i - 1] * i % MOD;
  for (int i = n; i >= 0; --i) {
    for (int j = 0; j <= i; ++j) {
      if (!(n & 1)) {
        if (j <= n / 2 && i - j <= n / 2)
          h[i] = ((long long)h[i] +
                  ((long long)f[n / 2][j][0] + f[n / 2][j][1]) *
                      ((long long)f[n / 2][i - j][0] + f[n / 2][i - j][1])) %
                 MOD;
      } else if (j <= n / 2 && i - j <= (n + 1) / 2) {
        long long a = f[n / 2][j][0] + f[n / 2][j][1],
                  b = g[(n + 1) / 2][i - j][0] + g[(n + 1) / 2][i - j][1];
        if (j) a += f[n / 2][j - 1][0];
        h[i] = ((long long)h[i] + a * b) % MOD;
      }
    }
    h[i] = (long long)h[i] * fac[n - i] % MOD;
    Tohka[i] = h[i];
    for (int j = i + 1; j <= n; ++j)
      Tohka[i] = (Tohka[i] - (long long)Tohka[j] * c[j][i] % MOD + MOD) % MOD;
  }
  cout << Tohka[k] << endl;
  fclose(stdin);
  fclose(stdout);
  return 0;
}
