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 int MOD = 1e9 + 7, N = 1000;
void add(int &a, int b) {
  if ((a += b) >= MOD) a -= MOD;
}
int bc[N + 1][N + 1], f[N + 1][N + 1][4], g[N + 1];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      bc[i][j] = j == 0 ? 1 : (bc[i - 1][j] + bc[i - 1][j - 1]) % MOD;
  f[0][0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      for (int k = 0; k < 4; ++k) {
        if ((k & 1) == 1) {
          if (j > 0 && i < n) {
            add(f[i][j][k], f[i - 1][j - 1][k >> 1]);
            add(f[i][j][k], f[i - 1][j - 1][k >> 1 | 2]);
          }
        } else {
          add(f[i][j][k], f[i - 1][j][k >> 1]);
          add(f[i][j][k], f[i - 1][j][k >> 1 | 2]);
          if (j > 0 && i > 1) add(f[i][j][k], f[i - 1][j - 1][k >> 1]);
        }
      }
  for (int i = 0; i <= n; ++i) {
    for (int k = 0; k < 4; k += 2) add(g[i], f[n][i][k]);
    for (int j = 1; j <= n - i; ++j) g[i] = 1LL * g[i] * j % MOD;
  }
  for (int i = n; i >= 0; --i)
    for (int j = i + 1; j <= n; ++j)
      add(g[i], MOD - 1LL * g[j] * bc[j][i] % MOD);
  cout << g[k] << '\n';
  return 0;
}
