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 = 1010, Mod = 1000000007;
int F[MaxN][MaxN][4], N, K, sum[MaxN], inv[MaxN];
inline void Exp(int &x, int y) {
  x += y;
  x >= Mod ? x -= Mod : x;
}
int main() {
  scanf("%d%d", &N, &K);
  F[0][0][1] = 1;
  for (int i = 0; i < N; i++)
    for (int j = 0; j <= i; j++)
      for (register int k = 0; k < 4; k++)
        if (F[i][j][k]) {
          Exp(F[i + 1][j][k >> 1], F[i][j][k]);
          Exp(F[i + 1][j + 1][(k >> 1) | 2], F[i][j][k]);
          if (!(k & 1)) Exp(F[i + 1][j + 1][k >> 1], F[i][j][k]);
        }
  long long tmp = 1;
  for (int j = N; j >= 0; j--, tmp = tmp * (N - j) % Mod) {
    for (register int k = 0; k < 2; k++) Exp(sum[j], tmp * F[N][j][k] % Mod);
  }
  inv[1] = 1;
  for (int i = 2; i <= N; i++)
    inv[i] = Mod - (long long)inv[Mod % i] * (Mod / i) % Mod;
  long long ans = 0, k = 1;
  for (int i = K; i <= N; i++, k = Mod - k) {
    ans = (ans + k * sum[i]) % Mod;
    k = k * (i + 1) % Mod * inv[i - K + 1] % Mod;
  }
  cout << ans << endl;
  return 0;
}
