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 = 1005, Mo = int(1e9) + 7;
int F[MAXN][MAXN][4], C[MAXN][MAXN], G[MAXN], Fac[MAXN], N, K;
void Update(int &a, int b) { a = (a + b) % Mo; }
int main() {
  scanf("%d%d", &N, &K);
  F[1][0][1] = 1;
  for (int i = 1; i <= N; i++) C[i][0] = C[0][i] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= i; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % Mo;
  Fac[0] = 1;
  for (int i = 1; i <= N; i++) Fac[i] = Fac[i - 1] * 1ll * i % Mo;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j < i; j++)
      for (int s = 0; s < 4; s++)
        if (F[i][j][s]) {
          if (!(s & 1)) Update(F[i + 1][j + 1][s >> 1], F[i][j][s]);
          if (i < N) Update(F[i + 1][j + 1][(s >> 1) | 2], F[i][j][s]);
          Update(F[i + 1][j][s >> 1], F[i][j][s]);
        }
  for (int j = 0; j <= N; j++) {
    for (int s = 0; s < 4; s++) G[j] = (G[j] + F[N + 1][j][s]) % Mo;
    G[j] = G[j] * 1ll * Fac[N - j] % Mo;
  }
  for (int j = N - 1; j >= 0; j--)
    for (int i = N; i > j; i--)
      G[j] = (G[j] - G[i] * 1ll * C[i][j] % Mo + Mo) % Mo;
  printf("%d\n", G[K]);
  return 0;
}
