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 = 2010;
const int mod = 1000000007;
long long dp[maxn][maxn][2][2];
long long cnk[maxn][maxn];
long long fact[maxn];
long long ans[maxn];
void addmod(long long& x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
void submod(long long& x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
int main(int argc, char** argv) {
  int n, k;
  scanf("%d %d", &n, &k);
  fact[0] = 1;
  for (int i = 1; i < maxn; ++i) {
    fact[i] = (1ll * fact[i - 1] * i) % mod;
  }
  cnk[0][0] = 1;
  for (int i = 1; i < maxn; ++i) {
    cnk[i][0] = 1;
    for (int j = 1; j < maxn; ++j) {
      cnk[i][j] = (cnk[i - 1][j - 1] + cnk[i - 1][j]) % mod;
    }
  }
  dp[0][0][0][0] = 1;
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= n; ++j)
      for (int x = 0; x < 2; ++x)
        for (int y = 0; y < 2; ++y) {
          addmod(dp[i + 1][j][y][0], dp[i][j][x][y]);
          if (!x && i) {
            addmod(dp[i + 1][j + 1][y][0], dp[i][j][x][y]);
          }
          if (i + 1 < n) {
            addmod(dp[i + 1][j + 1][y][1], dp[i][j][x][y]);
          }
        }
  for (int j = 0; j <= n; ++j) {
    for (int x = 0; x < 2; ++x) {
      for (int y = 0; y < 2; ++y) {
        addmod(ans[j], dp[n][j][x][y]);
      }
    }
    ans[j] = (1ll * ans[j] * fact[n - j]) % mod;
  }
  for (int i = n; i >= 0; --i) {
    for (int j = i + 1; j <= n; ++j) {
      submod(ans[i], (1ll * ans[j] * cnk[j][i]) % mod);
    }
  }
  cout << ans[k] << endl;
  return 0;
}
