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 mod = 1000000007;
int dp[1010][1010][2][2];
void Add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
long long ans[1010];
int C[1010][1010];
int main() {
  int n, K;
  scanf("%d%d", &n, &K);
  dp[0][0][1][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < n; j++) {
      for (int now = 0; now < 2; now++) {
        for (int next = 0; next < 2; next++) {
          if (!now) Add(dp[i][j + 1][next][0], dp[i - 1][j][now][next]);
          if (i < n) Add(dp[i][j + 1][next][1], dp[i - 1][j][now][next]);
          Add(dp[i][j][next][0], dp[i - 1][j][now][next]);
        }
      }
    }
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= i; j++)
      C[i][j] = j ? (C[i - 1][j] + C[i - 1][j - 1]) % mod : 1;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 2; j++) {
      for (int k = 0; k < 2; k++) {
        ans[i] += dp[n][i][j][k];
        ans[i] %= mod;
      }
    }
    for (int j = 2; j <= n - i; j++) ans[i] = ans[i] * j % mod;
  }
  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] % mod + mod) % mod;
    }
  }
  printf("%I64d\n", ans[K]);
  return 0;
}
