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;
int n, m, i, j, k;
int f[1005][1005][4], fac[1005], C[1005][1005], ans[1005];
void add(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
int main() {
  scanf("%d%d", &n, &m);
  fac[0] = 1;
  for (i = 1; i <= n; ++i) fac[i] = (long long)fac[i - 1] * i % 1000000007;
  C[0][0] = 1;
  for (i = 0; i <= n; ++i)
    for (j = 0; j <= i; ++j) {
      if (C[i][j] >= 1000000007) C[i][j] -= 1000000007;
      C[i + 1][j] += C[i][j];
      C[i + 1][j + 1] += C[i][j];
    }
  f[1][0][1] = 1;
  for (i = 1; i <= n; ++i)
    for (j = 0; j < i; ++j)
      for (k = 0; k < 4; ++k)
        if (f[i][j][k]) {
          add(f[i + 1][j][k >> 1], f[i][j][k]);
          if (!(k & 1)) add(f[i + 1][j + 1][k >> 1], f[i][j][k]);
          if (i != n) add(f[i + 1][j + 1][(k >> 1) + 2], f[i][j][k]);
        }
  for (i = 0; i <= n; ++i) {
    for (j = 0; j < 4; ++j) add(ans[i], f[n + 1][i][j]);
    ans[i] = (long long)ans[i] * fac[n - i] % 1000000007;
  }
  for (i = n; i >= 0; --i)
    for (j = i + 1; j <= n; ++j)
      add(ans[i], 1000000007 - (long long)ans[j] * C[j][i] % 1000000007);
  printf("%d\n", ans[m]);
}
