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;
int INP, AM, REACHEOF;
char BUF[(1 << 12) + 1], *inp = BUF;
const long double PI = acos((long double)-1.0);
const long long MOD = 1000000007LL;
long long f[1011][1011][5];
void add(long long &a, long long b) {
  if (b < 0) return;
  a += b;
  if (a >= MOD) a %= MOD;
}
int main() {
  f[1][0][0] = 1;
  for (int i = (2), _b = (1000); i <= _b; i++) {
    for (int k = (0), _b = (i); k <= _b; k++) {
      for (int x = 0, _a = (5); x < _a; x++) add(f[i][k][0], f[i - 1][k][x]);
      if (k > 1) f[i][k][1] = f[i - 1][k - 2][0];
      if (k > 0) {
        add(f[i][k][2], f[i - 1][k][1]);
        add(f[i][k][2], f[i - 1][k - 1][2]);
        add(f[i][k][2], f[i - 1][k][3]);
        add(f[i][k][2], f[i - 1][k - 1][4]);
      }
      if (k > 0) {
        add(f[i][k][3], f[i - 1][k][1]);
        add(f[i][k][3], f[i - 1][k][2]);
        add(f[i][k][3], f[i - 1][k - 1][3]);
        add(f[i][k][3], f[i - 1][k - 1][4]);
      }
      add(f[i][k][4], f[i - 1][k + 1][0] * (k + 1));
      add(f[i][k][4], f[i - 1][k + 1][1] * (k - 1));
      add(f[i][k][4], f[i - 1][k + 1][2] * k);
      add(f[i][k][4], f[i - 1][k + 1][3] * k);
      add(f[i][k][4], f[i - 1][k + 1][4] * (k + 1));
      add(f[i][k][4], f[i - 1][k][0] * (i - 2 - k));
      add(f[i][k][4], f[i - 1][k][1] * (i - 1 - k));
      add(f[i][k][4], f[i - 1][k][2] * (i - 2 - k));
      add(f[i][k][4], f[i - 1][k][3] * (i - 2 - k));
      add(f[i][k][4], f[i - 1][k][4] * (i - 3 - k));
    }
  }
  int k, n;
  while (cin >> n >> k) {
    long long res = 0;
    for (int t = 0, _a = (5); t < _a; t++) add(res, f[n][k][t]);
    cout << res << endl;
  }
  return 0;
}
