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 MOD = 1000000007;
void add(long long &a, long long b) {
  a += b;
  if (a < 0) a += MOD;
  if (a >= MOD) a -= MOD;
}
int n, k;
long long c[1020][1020], dp[1020][1020][4];
int main() {
  long long *f = new long long[1020];
  for (int i = 0; i < 1010; ++i) f[i] = 0;
  cin >> n >> k;
  for (int i = 0; i <= n; ++i) c[i][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
  dp[0][0][2] = 1;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      for (int q = 0; q < 4; ++q) {
        if (q & 1) add(dp[i + 1][j + 1][q / 2 + 2], dp[i][j][q]);
        if (i < n - 1) add(dp[i + 1][j + 1][q / 2], dp[i][j][q]);
        add(dp[i + 1][j][q / 2 + 2], dp[i][j][q]);
      }
  for (int i = 0; i <= n; ++i) {
    for (int q = 0; q < 4; ++q) add(f[i], dp[n][i][q]);
    for (int m = 2; m <= n - i; ++m) f[i] = f[i] * m % MOD;
  }
  for (int i = n - 1; i >= 0; --i)
    for (int j = i + 1; j <= n; ++j) {
      long long t = f[j] * c[j][i] % MOD;
      add(f[i], MOD - t);
    }
  cout << f[k] << endl;
}
