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 = 1000 + 5;
const int MOD = 1000000007;
int n, m, t, f[N][N][4], g[N], c[N][N];
inline void add(int& x, int y) {
  if ((x += y) > MOD) x -= MOD;
}
inline void sub(int& x, int y) {
  if ((x -= y) < 0) x += MOD;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      c[i][j] = j ? (c[i - 1][j - 1] + c[i - 1][j]) % MOD : 1;
  f[0][0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      for (int k = 0; k < 4; ++k) {
        if (i > 1 && j && !(k & 1)) add(f[i][j][k >> 1], f[i - 1][j - 1][k]);
        if (i < n && j) add(f[i][j][(k >> 1) | 2], f[i - 1][j - 1][k]);
        add(f[i][j][k >> 1], f[i - 1][j][k]);
      }
  for (int i = n, j = 1; i >= 0; j = j * (long long)(n - (--i)) % MOD)
    g[i] = (f[n][i][0] + f[n][i][1]) * (long long)j % MOD;
  for (int i = n; i >= m; --i)
    for (int j = n; j > i; --j) sub(g[i], g[j] * (long long)c[j][i] % MOD);
  printf("%d\n", g[m]);
  return 0;
}
