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 MAX_N = 1010, mod = 1e9 + 7;
int n, k, dp[MAX_N][MAX_N][4], binomial[MAX_N][MAX_N], f[MAX_N], fac[MAX_N];
int main() {
  scanf("%d%d", &n, &k);
  binomial[0][0] = 1;
  for (int i = fac[0] = 1; i <= n; i++) {
    fac[i] = 1LL * fac[i - 1] * i % mod;
    binomial[i][0] = binomial[i][i] = 1;
    for (int j = 1; j < i; j++)
      binomial[i][j] =
          (0LL + binomial[i - 1][j - 1] + binomial[i - 1][j]) % mod;
  }
  dp[0][0][1] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++)
      for (int S = 0; S < 4; S++) {
        dp[i][j][S >> 1] = (0LL + dp[i][j][S >> 1] + dp[i - 1][j][S]) % mod;
        if (~S & 1)
          dp[i][j + 1][S >> 1] =
              (0LL + dp[i][j + 1][S >> 1] + dp[i - 1][j][S]) % mod;
        if (i != n)
          dp[i][j + 1][(S >> 1) | 2] =
              (0LL + dp[i][j + 1][(S >> 1) | 2] + dp[i - 1][j][S]) % mod;
      }
  for (int i = 0; i <= n; i++)
    for (int stat = 0; stat < 4; stat++)
      f[i] = (0LL + f[i] + 1LL * dp[n][i][stat] * fac[n - i] % mod) % mod;
  int ans = 0;
  for (int i = k, opt = 1; i <= n; i++, opt = mod - opt)
    ans = (0LL + ans + 1LL * f[i] * binomial[i][k] % mod * opt % mod) % mod;
  printf("%d\n", ans);
  return 0;
}
