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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 1010, LOG = 20;
long long n, m, k, u, v, x, y, t, a, b;
long long dp[MAXN][MAXN];
long long C[MAXN][MAXN], F[MAXN];
long long res[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  dp[0][0] = dp[1][0] = 1;
  for (int i = 2; i <= n; i++) {
    dp[i][0] = 1;
    for (int j = 1; 2 * j <= i; j++)
      dp[i][j] = (dp[i - 1][j] + dp[i - 2][j - 1]) % mod;
  }
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= i; j++)
      res[i] = (res[i] + dp[n][j] * dp[n][i - j]) % mod;
  for (int i = 0; i <= n; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  }
  F[0] = 1;
  for (int i = 1; i <= n; i++) F[i] = F[i - 1] * i % mod;
  for (int i = 0; i <= n; i++) res[i] = res[i] * F[n - i] % mod;
  for (int i = n; i >= 0; i--) {
    for (int j = i + 1; j <= n; j++) res[i] = (res[i] - C[j][i] * res[j]) % mod;
  }
  if (res[k] < 0) res[k] += mod;
  cout << res[k] << '\n';
  return 0;
}
