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 = (1e9) + 7;
int f[1005][1005][6];
int C[1005][1005];
int jc[1005], ans[1005];
int n, m, i, j;
void init() {
  for (i = 1, jc[0] = 1; i <= n; i++) jc[i] = (long long)jc[i - 1] * i % Mod;
  for (i = 0; i <= n; i++) C[i][0] = 1;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= i; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % Mod;
}
int main() {
  scanf("%d%d", &n, &m);
  init();
  f[1][0][4] = 1;
  f[1][1][1] = 1;
  f[2][2][5] = 1;
  for (i = 1; i < n; i++)
    for (j = 0; j <= i; j++) {
      if (f[i][j][0]) {
        (f[i + 1][j + 1][0] += f[i][j][0]) %= Mod;
        (f[i + 1][j + 1][1] += f[i][j][0]) %= Mod;
        (f[i + 1][j][2] += f[i][j][0]) %= Mod;
        if (i < n - 1) (f[i + 2][j + 2][5] += f[i][j][0]) %= Mod;
      }
      if (f[i][j][1]) {
        (f[i + 1][j + 1][1] += f[i][j][1]) %= Mod;
        (f[i + 1][j][3] += f[i][j][1]) %= Mod;
      }
      if (f[i][j][2]) {
        (f[i + 1][j + 1][0] += f[i][j][2]) %= Mod;
        (f[i + 1][j + 1][1] += f[i][j][2]) %= Mod;
        (f[i + 1][j][4] += f[i][j][2]) %= Mod;
        if (i < n - 1) (f[i + 2][j + 2][5] += f[i][j][2]) %= Mod;
      }
      if (f[i][j][3]) {
        (f[i + 1][j + 1][1] += f[i][j][3]) %= Mod;
        (f[i + 1][j][4] += f[i][j][3]) %= Mod;
        if (i < n - 1) (f[i + 2][j + 2][5] += f[i][j][3]) %= Mod;
      }
      if (f[i][j][4]) {
        (f[i + 1][j + 1][0] += f[i][j][4]) %= Mod;
        (f[i + 1][j + 1][1] += f[i][j][4]) %= Mod;
        (f[i + 1][j][4] += f[i][j][4]) %= Mod;
        if (i < n - 1) (f[i + 2][j + 2][5] += f[i][j][4]) %= Mod;
      }
      if (f[i][j][5]) {
        (f[i + 1][j + 1][1] += f[i][j][5]) %= Mod;
        (f[i + 1][j][2] += f[i][j][5]) %= Mod;
        if (i < n - 1) (f[i + 2][j + 2][5] += f[i][j][5]) %= Mod;
      }
    }
  for (i = 0; i <= n; i++)
    ans[i] = (long long)((f[n][i][0] + f[n][i][2]) % Mod +
                         ((f[n][i][3] + f[n][i][4]) % Mod + f[n][i][5]) % Mod) *
             jc[n - i] % Mod;
  for (i = n; i >= 0; i--)
    for (j = i + 1; j <= n; j++)
      (ans[i] -= (long long)ans[j] * C[j][i] % Mod - Mod) %= Mod;
  printf("%d\n", ans[m]);
  return 0;
}
