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 N = 1005;
const int mod = 1e9 + 7;
int n, m;
int c[N][N], fac[N], f[N][N][2][2], dp[N], g[N];
void Init() {
  scanf("%d%d", &n, &m);
  c[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
      if (c[i][j] >= mod) c[i][j] -= mod;
    }
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
}
inline void Add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  Init();
  f[1][0][1][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      for (int a = 0; a < 2; a++) {
        for (int b = 0; b < 2; b++) {
          int t = f[i][j][a][b];
          Add(f[i + 1][j][b][0], t);
          Add(f[i + 1][j + 1][b][1], t);
          if (a == 0) Add(f[i + 1][j + 1][b][0], t);
        }
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 2; j++) {
      Add(dp[i], f[n + 1][i][j][0]);
    }
  }
  for (int i = n; i >= 0; i--) {
    g[i] = 1LL * dp[i] * fac[n - i] % mod;
    for (int j = i + 1; j <= n; j++) {
      g[i] -= 1LL * c[j][i] * g[j] % mod;
      if (g[i] < 0) g[i] += mod;
    }
  }
  printf("%d\n", g[m]);
  return 0;
}
