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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
using namespace std;
const int N = 2005, INF = 0x3f3f3f3f;
const long long MOD = 1000000007;
int n, k;
long long f[N][N][2][2], C[N][N], g[N], fac[N] = {1};
void init() {
  for (register int i = 0; i <= n; i++) {
    f[i][0][0][0] = 1;
    for (register int j = 0; j <= n; j++) {
      if (!i && j) continue;
      for (register int a = 0; a < 2; a++)
        for (register int b = 0; b < 2; b++) {
          (f[i + 1][j][b][0] += f[i][j][a][b]) %= MOD;
          if (!a && i) (f[i + 1][j + 1][b][0] += f[i][j][a][b]) %= MOD;
          (f[i + 1][j + 1][b][1] += f[i][j][a][b]) %= MOD;
        }
    }
  }
  for (register int i = 1; i <= n; i++) {
    C[i][0] = C[i][i] = 1;
    for (register int j = 1; j < i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
  }
  for (register int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % MOD;
}
void solve() {
  for (register int i = n; i >= 0; i--) {
    long long Left = (f[n][i][1][0] + f[n][i][0][0]) * fac[n - i] % MOD;
    long long Right = 0;
    for (register int j = n; j > i; j--) Right = (Right + C[j][i] * g[j]) % MOD;
    g[i] = ((Left - Right) % MOD + MOD) % MOD;
  }
}
signed main() {
  scanf("%d%d", &n, &k);
  init();
  solve();
  printf("%lld\n", g[k]);
  return 0;
}
