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 long long modc = 1e9 + 7;
const int nmax = 1000 + 10;
long long f[nmax][nmax][4];
long long F[nmax];
long long c[nmax][nmax];
int n, k;
inline void addmod(long long &x, long long y) { x = (x + y) % modc; }
inline void mulmod(long long &x, long long y) { x = (x * y) % modc; }
void pascal() {
  c[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      addmod(c[i][j], c[i - 1][j]);
      addmod(c[i][j], c[i - 1][j - 1]);
    }
  }
}
void debug() {
  for (int i = 1; i <= n; i++) {
    cerr << i << " ";
    for (int j = 0; j <= n; j++) {
      cerr << "(";
      for (int z = 0; z < 4; z++) {
        cerr << f[i][j][z] << " ";
      }
      cerr << ") ";
    }
    cerr << '\n';
  }
  cerr << '\n';
}
void debug2() {
  for (int i = 0; i <= n; i++) {
    cerr << F[i] << " ";
  }
  cerr << '\n';
}
int main() {
  cin >> n >> k;
  if (n == 1) {
    cout << (k ? 0 : 1) << '\n';
    return 0;
  }
  pascal();
  f[1][0][0] = 1;
  f[1][1][2] = 1;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      for (int z = 0; z < 4; z++) {
        addmod(f[i + 1][j][z / 2], f[i][j][z]);
        if ((z & 1) == 0) {
          addmod(f[i + 1][j + 1][z / 2], f[i][j][z]);
        }
        if (i < n - 1) {
          addmod(f[i + 1][j + 1][z / 2 + 2], f[i][j][z]);
        }
      }
    }
  }
  for (int j = 0; j <= n; j++) {
    for (int z = 0; z < 4; z++) addmod(F[j], f[n][j][z]);
    for (int a = 2; a <= n - j; a++) mulmod(F[j], a);
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i + 1; j <= n; j++) {
      long long tmp = F[j];
      mulmod(tmp, c[j][i]);
      addmod(F[i], modc - tmp);
    }
  }
  cout << F[k] << '\n';
  return 0;
}
