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 void SetIO(string Name) {
  string Input = Name + ".in", Output = Name + ".out";
  freopen(Input.c_str(), "r", stdin), freopen(Output.c_str(), "w", stdout);
}
const int N = 1010, Mod = 1000000007;
int n, m;
long long C[N][N], Dp[N][N][4], Ans[N];
inline void Input() { scanf("%d%d", &n, &m); }
inline void Plus(long long &A, long long B) { A = (A + B) % Mod; }
inline void Mul(long long &A, int B) { A = (A * B) % Mod; }
inline void Solve() {
  for (int i = (0); i <= (n); i++) C[i][0] = 1;
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (i); j++)
      Plus(C[i][j], C[i - 1][j - 1] + C[i - 1][j]);
  Dp[0][0][2] = 1;
  for (int i = (1); i <= (n); i++)
    for (int j = 0; j < (i); j++)
      for (int S = 0; S < (4); S++) {
        if (S & 1) Plus(Dp[i][j + 1][(S >> 1) + 2], Dp[i - 1][j][S]);
        if (i < n) Plus(Dp[i][j + 1][S >> 1], Dp[i - 1][j][S]);
        Plus(Dp[i][j][(S >> 1) + 2], Dp[i - 1][j][S]);
      }
  for (int i = (0); i <= (n); i++) {
    for (int S = 0; S < (4); S++) Plus(Ans[i], Dp[n][i][S]);
    for (int j = (1); j <= (n - i); j++) Mul(Ans[i], j);
  }
  for (int i = (n)-1; i >= (0); i--)
    for (int j = (i + 1); j <= (n); j++)
      Plus(Ans[i], Mod - C[j][i] * Ans[j] % Mod);
  cout << Ans[m] << endl;
}
int main() {
  Input();
  Solve();
  return 0;
}
