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 N = 1005;
const int mod = 1e9 + 7;
int n, K, F[N];
int dp[N][N][4];
int C[N][N];
int fact[N];
int solve(int i, int baki, int f) {
  if (baki < 0 or i > n + 1) return 0;
  if (i > n) return baki == 0;
  if (dp[i][baki][f] != -1) return dp[i][baki][f];
  int ways = 0;
  if (f == 0) {
    ways = solve(i + 2, baki - 2, 1);
    ways += solve(i + 1, baki, 0);
    ways %= mod;
    ways += solve(i + 1, baki - 1, 0);
    ways %= mod;
    if (i != n) ways += solve(i + 1, baki - 1, 2);
    ways %= mod;
  } else if (f == 1) {
    ways = solve(i + 2, baki - 2, 1);
    ways += solve(i + 1, baki, 0);
    ways %= mod;
    if (i != n) ways += solve(i + 1, baki - 1, 2);
    ways %= mod;
  } else {
    ways = solve(i + 1, baki, 1);
    if (i != n) ways += solve(i + 1, baki - 1, 2);
    ways %= mod;
  }
  return dp[i][baki][f] = ways;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> K;
  memset(dp, -1, sizeof dp);
  C[0][0] = 1;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    C[i][0] = 1;
    fact[i] = 1ll * fact[i - 1] * i % mod;
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
  for (int i = n; i >= 0; i--) {
    F[i] = 1ll * solve(1, i, 1) * fact[n - i] % mod;
    for (int j = i + 1; j <= n; j++) {
      F[i] -= 1ll * F[j] * C[j][i] % mod;
      if (F[i] < 0) F[i] += mod;
    }
  }
  cout << F[K] << endl;
  return 0;
}
