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 N = 1003;
const int mod = 1000000007;
int n, k, tag;
int fac[N], inv[N], dp[N][N][2][2], f[N];
int quickpow(int x, int y) {
  int s = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) s = 1ll * s * x % mod;
  return s;
}
int C(int x, int y) {
  if (x < y) return 0;
  return 1ll * fac[x] * inv[y] % mod * 1ll * inv[x - y] % mod;
}
void add(int &x, int &y) {
  x += y;
  x = (x > mod) ? x - mod : x;
}
int main() {
  scanf("%d%d", &n, &k);
  inv[0] = fac[0] = inv[1] = fac[1] = 1;
  for (int i = 2; i <= n; i++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
    inv[i] = quickpow(fac[i], mod - 2);
  }
  dp[0][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++)
      for (int a = 0; a < 2; a++)
        for (int b = 0; b < 2; b++) {
          if (!a) add(dp[i][j + 1][b][0], dp[i - 1][j][a][b]);
          if (i < n) add(dp[i][j + 1][b][1], dp[i - 1][j][a][b]);
          add(dp[i][j][b][0], dp[i - 1][j][a][b]);
        }
  for (int i = 0; i <= n; i++) {
    for (int a = 0; a < 2; a++)
      for (int b = 0; b < 2; b++) add(f[i], dp[n][i][a][b]);
    f[i] = 1ll * f[i] * fac[n - i] % mod;
  }
  tag = -1;
  for (int i = k + 1; i <= n; i++) {
    f[k] += 1ll * tag * f[i] * C(i, k) % mod;
    f[k] %= mod;
    tag *= -1;
  }
  printf("%d\n", (f[k] + mod) % mod);
  return 0;
}
