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 long long Mod = 1e9 + 7;
int n, k, f[1010][1010][2][3];
long long C[1010][1010], ans[1010], mul[1010], Ans;
void get_C(int N) {
  for (int i = 0; i <= N; i++) C[i][0] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % Mod;
}
int main() {
  scanf("%d%d", &n, &k);
  get_C(n);
  mul[0] = 1;
  f[0][0][1][0] = 1;
  for (int i = 1; i <= n; i++) mul[i] = mul[i - 1] * i % Mod;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++)
      for (int p = 0; p <= 1; p++)
        for (int loc = 0; loc <= 2; loc++)
          if (f[i][j][p][loc] != 0) {
            int np = (loc == 2) ? 1 : 0, now;
            now = f[i][j][p][loc];
            f[i + 1][j][np][0] = (f[i + 1][j][np][0] + now) % Mod;
            if (p == 0)
              f[i + 1][j + 1][np][1] = (f[i + 1][j + 1][np][1] + now) % Mod;
            if (i + 1 < n)
              f[i + 1][j + 1][np][2] = (f[i + 1][j + 1][np][2] + now) % Mod;
          }
  for (int i = 0; i <= n; i++)
    for (int p = 0; p <= 1; p++)
      for (int loc = 0; loc <= 2; loc++)
        ans[i] = (ans[i] + f[n][i][p][loc]) % Mod;
  for (int i = 0; i <= n; i++) ans[i] = (ans[i] * mul[n - i]) % Mod;
  for (int i = k, dlt = 1; i <= n; i++, dlt = -dlt)
    Ans = (Ans + dlt * ans[i] * C[i][k] % Mod) % Mod;
  Ans = (Ans + Mod) % Mod;
  printf("%I64d\n", Ans);
  return 0;
}
