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;
inline int read() {
  int res = 0;
  int neg;
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9' || ch == '-') {
      if (ch == '-')
        neg = -1;
      else
        neg = 1, res = ch - '0';
      break;
    }
  }
  while (true) {
    char ch = getchar();
    if (ch >= '0' && ch <= '9')
      res *= 10, res += ch - '0';
    else
      break;
  }
  return res * neg;
}
const int maxn = 1020;
const int maxm = 1020;
const int MOd = 1e9 + 7;
int a, b;
long long dn[maxn][maxn][2][2], sum[maxn], fakt[maxn], comb[maxn][maxn];
int main() {
  scanf("%d %d", &a, &b);
  dn[0][0][0][0] = 1;
  comb[0][0] = fakt[0] = 1;
  for (int i = 1; i <= a; i++) comb[i][0] = 1, fakt[i] = fakt[i - 1] * i % MOd;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= i; j++)
      comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % MOd;
  for (int i = 0; i <= a; i++)
    for (int j = 0; j <= i; j++)
      for (int b1 = 0; b1 <= 1; b1++)
        for (int b2 = 0; b2 <= 1; b2++) {
          if (!b2 && i > 0)
            dn[i + 1][j + 1][0][b1] += dn[i][j][b1][b2],
                dn[i + 1][j + 1][0][b1] %= MOd;
          if (i + 1 < a)
            dn[i + 1][j + 1][1][b1] += dn[i][j][b1][b2],
                dn[i + 1][j + 1][1][b1] %= MOd;
          dn[i + 1][j][0][b1] += dn[i][j][b1][b2];
          dn[i + 1][j][0][b1] %= MOd;
        }
  for (int i = 0; i <= a; i++) {
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++) sum[i] += dn[a][i][j][k];
    sum[i] *= fakt[a - i];
    sum[i] %= MOd;
  }
  for (int i = a - 1; i >= 0; i--)
    for (int j = i + 1; j <= a; j++) {
      sum[i] -= sum[j] * comb[j][i] % MOd;
      if (sum[i] < 0) sum[i] += MOd;
    }
  cout << sum[b] << endl;
  return 0;
}
