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;
using ll = long long;
using ull = unsigned long long;
const int N = 1005;
const ll mod = 1e9 + 7;
const int base = 311;
ll m, n, k, t, T, ans, u, v, tong;
ll a[N], b[N], d[N], f[N][N][4];
string s[105];
vector<ll> kq;
ll C(ll u, ll v) {
  if (u > v) return 0;
  return a[v] * b[u] % mod * b[v - u] % mod;
}
ll pw(ll k, ll n) {
  ll total = 1;
  for (; n; n >>= 1) {
    if (n & 1) total = (total * k) % mod;
    k = (k * k) % mod;
  }
  return total;
}
void sol() {
  cin >> n >> k;
  a[0] = b[0] = 1;
  for (int i = 1; i <= n; i++) {
    a[i] = a[i - 1] * i % mod;
    b[i] = pw(i, mod - 2);
    b[i] = b[i - 1] * b[i] % mod;
  }
  f[0][0][1] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      for (int z = 0; z < 4; z++) {
        if (i < n)
          f[i][j + 1][z / 2 + 2] =
              (f[i][j + 1][z / 2 + 2] + f[i - 1][j][z]) % mod;
        f[i][j][z / 2] = (f[i][j][z / 2] + f[i - 1][j][z]) % mod;
        if (z % 2 == 0)
          f[i][j + 1][z / 2] = (f[i][j + 1][z / 2] + f[i - 1][j][z]) % mod;
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int z = 0; z < 4; z++)
      d[i] = (d[i] + f[n][i][z] * a[n - i] % mod) % mod;
  }
  for (int i = n - 1; i >= k; i--) {
    for (int j = i + 1; j <= n; j++) {
      d[i] = (d[i] - d[j] * C(i, j) % mod + mod) % mod;
    }
  }
  cout << d[k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("LAZY"
            ".inp",
            "r")) {
    freopen(
        "LAZY"
        ".inp",
        "r", stdin);
    freopen(
        "LAZY"
        ".out",
        "w", stdout);
  }
  sol();
}
