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 Input() {
  char c = getchar();
  int x = 0, z = 1;
  for (; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  return x * z;
}
const int mod(1e9 + 7);
const int maxn(1005);
int n, m, f[maxn][maxn], g[maxn], c[maxn][maxn], fac[maxn];
void Upd(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  n = Input(), m = Input(), fac[0] = f[0][0] = c[0][0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = 1LL * fac[i - 1] * i % mod;
  for (int i = 1, j; i <= n; ++i)
    for (c[i][0] = j = 1; j <= i; ++j)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= i; ++j) {
      f[i][j] = f[i - 1][j];
      if (j) Upd(f[i][j], f[max(0, i - 2)][j - 1]);
    }
  int num = n - 1;
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= i; ++j)
      Upd(g[i], 1LL * f[num][j] * f[num][i - j] % mod);
    g[i] = 1LL * g[i] * fac[n - i] % mod;
  }
  for (int i = n - 1; ~i; --i)
    for (int j = i + 1; j <= n; ++j)
      Upd(g[i], mod - 1LL * g[j] * c[j][i] % mod);
  printf("%d\n", g[m]);
  return 0;
}
