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 = (int)1e9 + 7;
const int N = 1003;
inline void iadd(int& x, int y) {
  x += y;
  x -= x >= mod ? mod : 0;
}
inline int add(int x, int y) {
  iadd(x, y);
  return x;
}
int dp[N][N][2][2];
int dp2[N][N];
int n, k;
void solve() {
  cin >> n >> k;
  dp2[1][0] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      iadd(dp[i + 1][j][0][0], dp2[i][j] * 1ll * (i - j - 1) % mod);
      iadd(dp[i + 1][j + 2][1][1], dp2[i][j]);
      if (j) iadd(dp[i + 1][j - 1][0][0], dp2[i][j] * 1ll * j % mod);
      iadd(dp2[i + 1][j], dp2[i][j]);
      for (int f1 = 0; f1 < 2; f1++) {
        for (int f2 = 0; f2 < 2; f2++) {
          if (!dp[i][j][f1][f2]) continue;
          int pure = j - f1 - f2;
          int pure2 = i - j - (1 - f1) - (1 - f2);
          if (pure < 0 || pure2 < 0) continue;
          iadd(dp2[i + 1][j], dp[i][j][f1][f2]);
          iadd(dp[i + 1][j][0][0], dp[i][j][f1][f2] * 1ll * pure2 % mod);
          if (j)
            iadd(dp[i + 1][j - 1][0][0], dp[i][j][f1][f2] * 1ll * pure % mod);
          if (f1)
            iadd(dp[i + 1][j][0][1], dp[i][j][f1][f2]);
          else
            iadd(dp[i + 1][j + 1][0][1], dp[i][j][f1][f2]);
          if (f2)
            iadd(dp[i + 1][j][1][0], dp[i][j][f1][f2]);
          else
            iadd(dp[i + 1][j + 1][1][0], dp[i][j][f1][f2]);
        }
      }
    }
  }
  cout << add(add(add(dp[n][k][0][0], dp[n][k][0][1]),
                  add(dp[n][k][1][0], dp[n][k][1][1])),
              dp2[n][k])
       << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T = 1;
  while (T--) solve();
  return 0;
}
