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;
int dp[1010][1010][2][2][2];
void add(int &x, int o) {
  x += o;
  if (x >= mod) x -= mod;
}
int main() {
  int n, k, z, c0, c1, ans = 0;
  scanf("%d%d", &n, &k);
  dp[1][0][1][0][0] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j <= i; j++)
      for (int a = 0; a <= 1; a++)
        for (int b = 0; b <= 1; b++)
          for (int c = 0; c <= 1; c++) {
            z = dp[i][j][a][b][c];
            c1 = j;
            c0 = i - j;
            add(dp[i + 1][j][1][0][0], z);
            add(dp[i + 1][j + 1 - b + a][0][a][1], z);
            if (b)
              c1--;
            else
              c0--;
            if (!a) {
              add(dp[i + 1][j + 1 - c][0][1][0], z);
              if (c)
                c1--;
              else
                c0--;
            }
            if (c0) add(dp[i + 1][j][0][0][0], (1LL * z * c0) % mod);
            if (c1) add(dp[i + 1][j - 1][0][0][0], (1LL * z * c1) % mod);
          }
  for (int a = 0; a <= 1; a++)
    for (int b = 0; b <= 1; b++)
      for (int c = 0; c <= 1; c++) add(ans, dp[n][k][a][b][c]);
  printf("%d", ans);
  return 0;
}
