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>
const int mo = 1e9 + 7;
const int maxn = 1e3 + 5;
using namespace std;
int n, m, f[maxn][maxn][2][2];
int fac[maxn], inv[maxn], ans;
void add(int &x, int y) {
  if (y > mo) y -= mo;
  x = (x + y) % mo;
}
int ksm(int a, int b) {
  int sum = 1;
  for (; b; b >>= 1, a = 1LL * a * a % mo)
    if (b & 1) sum = 1LL * sum * a % mo;
  return sum;
}
int C(int a, int b) { return 1LL * fac[a] * inv[b] % mo * inv[a - b] % mo; }
int main() {
  scanf("%d %d", &n, &m);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1LL * fac[i - 1] * i % mo;
  inv[n] = ksm(fac[n], mo - 2);
  for (int i = n; i >= 1; i--) inv[i - 1] = 1LL * inv[i] * i % mo;
  f[1][0][0][0] = 1;
  f[1][1][0][1] = 1;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      if (j) add(f[i][j][0][0], f[i - 1][j - 1][0][0]);
      if (j) add(f[i][j][1][0], f[i - 1][j - 1][0][1]);
      if (j) add(f[i][j][0][1], f[i - 1][j - 1][0][0] + f[i - 1][j - 1][1][0]);
      if (j) add(f[i][j][1][1], f[i - 1][j - 1][0][1] + f[i - 1][j - 1][1][1]);
      add(f[i][j][0][0], f[i - 1][j][0][0] + f[i - 1][j][1][0]);
      add(f[i][j][1][0], f[i - 1][j][0][1] + f[i - 1][j][1][1]);
    }
  }
  for (int i = m; i <= n; i++) {
    int x = 1LL * (f[n][i][0][0] + f[n][i][1][0]) % mo * fac[n - i] % mo *
            C(i, m) % mo;
    if ((i - m) & 1)
      ans = (ans - x + mo) % mo;
    else
      ans = (ans + x) % mo;
  }
  printf("%d\n", ans);
  return 0;
}
