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 maxn = 1024;
const int MOD = 1000000007;
long long c[maxn][maxn], dp[maxn][maxn][4], ans[maxn];
void update(long long &a, long long b) { a = (a + b) % MOD; }
int main() {
  int n, K;
  cin >> n >> K;
  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;
  dp[0][0][2] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      for (int k = 0; k < 4; k++) {
        if (k & 1) update(dp[i][j + 1][(k >> 1) + 2], dp[i - 1][j][k]);
        if (i < n) update(dp[i][j + 1][k >> 1], dp[i - 1][j][k]);
        update(dp[i][j][(k >> 1) + 2], dp[i - 1][j][k]);
      }
    }
  }
  for (int j = 0; j <= n; j++) {
    for (int k = 0; k < 4; k++) update(ans[j], dp[n][j][k]);
    for (int k = 1; k <= n - j; k++) ans[j] = ans[j] * k % MOD;
  }
  for (int i = n - 1; i >= 0; i--)
    for (int j = i + 1; j <= n; j++)
      update(ans[i], MOD - c[j][i] * ans[j] % MOD);
  cout << ans[K] << endl;
  return 0;
}
