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 f[1010][1010][2][2], n, K;
const int p = 1000000007;
long long g[1010], fac[1010], c[1010][1010], ans[1010];
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
void add(int &x, int y) {
  x += y;
  if (x >= p) x -= p;
}
int main() {
  n = read();
  K = read();
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * (long long)i) % p;
  c[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % p;
  }
  f[0][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < n; j++)
      for (int now = 0; now <= 1; now++)
        for (int next = 0; next <= 1; next++) {
          if (!now) add(f[i][j + 1][next][0], f[i - 1][j][now][next]);
          if (i < n) add(f[i][j + 1][next][1], f[i - 1][j][now][next]);
          add(f[i][j][next][0], f[i - 1][j][now][next]);
        }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= 1; j++) {
      for (int k = 0; k <= 1; k++) {
        ans[i] += f[n][i][j][k];
        ans[i] %= p;
      }
    }
    ans[i] = ans[i] * fac[n - i] % p;
  }
  for (int i = n; i >= 0; i--) {
    for (int j = i + 1; j <= n; j++) {
      ans[i] -= ans[j] * c[j][i];
      ans[i] = (ans[i] % p + p) % p;
    }
  }
  cout << ans[K] << endl;
}
