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 MaxN = 1010;
const int MOD = 1000000007;
int n, k;
int C[MaxN][MaxN];
int F[MaxN][MaxN][4];
int p[MaxN];
void add(int &x, int y) { x = (x + y) % MOD; }
int main() {
  cin >> n >> k;
  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)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
  F[0][0][1] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      for (int s = 0; s < 4; ++s) {
        add(F[i][j][s / 2], F[i - 1][j][s]);
        add(F[i][j + 1][s / 2 + 2], F[i - 1][j][s]);
        if (s % 2 == 0) add(F[i][j + 1][s / 2], F[i - 1][j][s]);
      }
  for (int i = 0; i <= n; ++i) {
    add(p[i], F[n][i][0]);
    add(p[i], F[n][i][1]);
    for (int j = 1; j <= n - i; ++j) p[i] = (long long)p[i] * j % MOD;
  }
  for (int i = n; i >= 0; --i)
    for (int j = i + 1; j <= n; ++j)
      add(p[i], MOD - (long long)p[j] * C[j][i] % MOD);
  cout << p[k] << endl;
  return 0;
}
