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 Mod = 1e9 + 7;
const long long INF = (long long)(1e18) + 5;
const long long N = 1e3 + 5, M = 18;
int n, k;
int dp[N][N];
long long fact[N];
long long fp(long long x, long long y, long long m = Mod) {
  long long res = 1;
  for (int i = 0; (1LL << i) <= y; i++, x = x * x % m)
    if (y & 1LL << i) res = res * x % m;
  return res;
}
long long C(int n, int r) {
  return fact[n] * fp(fact[n - r] * fact[r] % Mod, Mod - 2) % Mod;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % Mod;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i; j++) {
      dp[i][j] =
          (dp[i - 1][j] - (i - 1 && j ? dp[i - 2][j - 1] - Mod : 0)) % Mod;
      if (i >= 2 && j >= 2)
        dp[i][j] = (dp[i][j] + (long long)dp[i - 2][j - 2] -
                    (i - 2 && j - 2 ? dp[i - 3][j - 3] - Mod : 0)) %
                   Mod;
      if (i - 2 >= 0 && j - 1 >= 0)
        dp[i][j] = (dp[i][j] + dp[i - 2][j - 1] * 2 % Mod) % Mod;
      if (i - 1 >= 0 && j - 1 >= 0)
        dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % Mod;
    }
  for (int i = n; i > 0; i--)
    dp[n][i] = (dp[n][i] - dp[n - 1][i - 1] + Mod) % Mod;
  long long ans = dp[n][k] * fact[n - k] % Mod;
  for (int i = k + 1; i <= n; i++)
    ans += (i % 2 == k % 2 ? 1 : -1) * dp[n][i] * fact[n - i] % Mod * C(i, k) %
           Mod;
  cout << (ans % Mod + Mod) % Mod << endl;
  return 0;
}
