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;
template <class T>
inline int size(const T& c) {
  return c.size();
}
using namespace std;
const int maxn = 1000 + 10;
const int cmod = 1000000007;
long long F[maxn][maxn][5];
int s1, s0;
int n, k;
int main() {
  cin >> n >> k;
  F[1][0][4] = 1;
  for (int i = (1), _b = (n - 1); i <= _b; i++)
    for (int j = (0), _b = (i); j <= _b; j++)
      for (int t = (0), _b = (4); t <= _b; t++) {
        if (F[i][j][t] == 0) continue;
        if (t == 4) {
          s1 = j;
          s0 = i - j - 1;
          F[i + 1][j][0 + 0] = (F[i + 1][j][0 + 0] + F[i][j][t] * s0) % cmod;
          F[i + 1][j - 1][0 + 0] =
              (F[i + 1][j - 1][0 + 0] + F[i][j][t] * s1) % cmod;
          F[i + 1][j + 2][2 + 1] = (F[i + 1][j + 2][2 + 1] + F[i][j][t]) % cmod;
          F[i + 1][j][4] = (F[i + 1][j][4] + F[i][j][t]) % cmod;
          continue;
        }
        int bstn = t / 2;
        int bn = t % 2;
        if (bstn == 1)
          s1 = j - 1, s0 = i - j;
        else
          s1 = j, s0 = i - j - 1;
        if (bn == 1)
          s1--;
        else
          s0--;
        F[i + 1][j][0] = (F[i + 1][j][0] + F[i][j][t] * s0) % cmod;
        F[i + 1][j - 1][0] = (F[i + 1][j - 1][0] + F[i][j][t] * s1) % cmod;
        if (bstn == 0)
          F[i + 1][j + 1][0 + 1] = (F[i + 1][j + 1][0 + 1] + F[i][j][t]) % cmod;
        else
          F[i + 1][j][0 + 1] = (F[i + 1][j][0 + 1] + F[i][j][t]) % cmod;
        if (bn == 0)
          F[i + 1][j + 1][2 + 0] = (F[i + 1][j + 1][2 + 0] + F[i][j][t]) % cmod;
        else
          F[i + 1][j][2 + 0] = (F[i + 1][j][2 + 0] + F[i][j][t]) % cmod;
        F[i + 1][j][4] = (F[i + 1][j][4] + F[i][j][t]) % cmod;
      }
  int res = 0;
  for (int t = (0), _b = (4); t <= _b; t++) res = (res + F[n][k][t]) % cmod;
  cout << res << endl;
  fclose(stdin);
  fclose(stdout);
  return 0;
}
