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 mod = 1e9 + 7;
long long f[1005][1005][5];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  f[1][0][4] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      for (int t = 0; t <= 3; t++) {
        int s1, s2;
        if (t == 0) s1 = i - j - 2, s2 = j;
        if (t == 1 || t == 2) s1 = i - j - 1, s2 = j - 1;
        if (t == 3) s1 = i - j, s2 = j - 2;
        f[i + 1][j][0] = (f[i + 1][j][0] + (f[i][j][t] * s1)) % mod;
        f[i + 1][j - 1][0] = (f[i + 1][j - 1][0] + (f[i][j][t] * s2)) % mod;
        f[i + 1][j + ((t & 1) ? 0 : 1)][2] =
            (f[i + 1][j + ((t & 1) ? 0 : 1)][2] + (f[i][j][t])) % mod;
        f[i + 1][j + (t < 2 ? 1 : 0)][1] =
            (f[i + 1][j + (t < 2 ? 1 : 0)][1] + (f[i][j][t])) % mod;
        f[i + 1][j][4] = (f[i + 1][j][4] + (f[i][j][t])) % mod;
      }
      f[i + 1][j][0] = (f[i + 1][j][0] + (f[i][j][4] * (i - j - 1))) % mod;
      f[i + 1][j - 1][0] = (f[i + 1][j - 1][0] + (f[i][j][4] * j)) % mod;
      f[i + 1][j + 2][3] = (f[i + 1][j + 2][3] + (f[i][j][4])) % mod;
      f[i + 1][j][4] = (f[i + 1][j][4] + (f[i][j][4])) % mod;
    }
  }
  int ans = 0;
  for (int i = 0; i <= 4; i++) ans = (ans + (f[n][k][i])) % mod;
  printf("%d", ans);
}
