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 mul(int x, int y) { return 1LL * x * y % 1000000007; }
void add(int &x, int y) { x = (x + y) % 1000000007; }
int f[1024][1024][2][2];
int jc[1024];
int h[1024];
int c[1024][1024];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  jc[0] = 1;
  for (int i = 1; i <= n; i++) jc[i] = mul(jc[i - 1], i);
  for (int i = 1; i <= n; i++) {
    c[i][0] = 1, c[i][i] = 1;
    for (int j = 1; j < i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  }
  memset(h, 0, sizeof(h));
  memset(f, 0, sizeof(f));
  f[1][0][0][0] = 1;
  f[1][1][0][1] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i; j++)
      for (int x1 = 0; x1 < 2; x1++)
        for (int x2 = 0; x2 < 2; x2++) {
          int p = f[i][j][x1][x2];
          if (x1 == 0 && x2 == 0) {
            add(f[i + 1][j + 1][0][0], p);
            add(f[i + 1][j + 1][0][1], p);
            add(f[i + 1][j][0][0], p);
          }
          if (x1 == 0 && x2 == 1) {
            add(f[i + 1][j + 1][1][0], p);
            add(f[i + 1][j + 1][1][1], p);
            add(f[i + 1][j][1][0], p);
          }
          if (x1 == 1 && x2 == 0) {
            add(f[i + 1][j + 1][0][1], p);
            add(f[i + 1][j][0][0], p);
          }
          if (x1 == 1 && x2 == 1) {
            add(f[i + 1][j + 1][1][1], p);
            add(f[i + 1][j][1][0], p);
          }
        }
  for (int i = n; ~i; i--) {
    h[i] = (f[n][i][0][0] + f[n][i][1][0]) % 1000000007;
    h[i] = mul(h[i], jc[n - i]), h[i] %= 1000000007;
    for (int j = i + 1; j <= n; j++) {
      h[i] = h[i] - mul(h[j], c[j][i]);
      h[i] = (h[i] + 1000000007) % 1000000007;
    }
  }
  printf("%d\n", h[k]);
  scanf("%d", &n);
  return 0;
}
