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 inf = 0x20202020;
const int mod = 1000000007;
template <class T>
inline void read(T &first) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (first = 0; c > 32; c = getchar()) first = first * 10 + c - '0';
  if (fu) first = -first;
};
template <class T>
inline void read(T &first, T &second) {
  read(first);
  read(second);
}
template <class T>
inline void read(T &first, T &second, T &z) {
  read(first);
  read(second);
  read(z);
}
template <class T>
inline void read(T &first, T &second, T &z, T &q) {
  read(first);
  read(second);
  read(z);
  read(q);
}
const int DX[] = {1, 0, -1, 0}, DY[] = {0, 1, 0, -1};
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const int N = 1111, M = 111111;
int l, m, n, t, C;
long long f[N][N][2][2], g[N], jc[N], inv[N];
void add(long long &first, long long second) { first = (first + second) % mod; }
int main() {
  scanf("%d%d", &n, &m);
  jc[0] = inv[0] = 1;
  for (int i = (1); i <= (n); ++i)
    jc[i] = jc[i - 1] * i % mod, inv[i] = powmod(jc[i], mod - 2);
  f[0][0][0][1] = 1;
  for (int i = (1); i <= (n); ++i)
    for (int j = (0); j <= (i); ++j)
      for (int k = (0); k <= (1); ++k) {
        if (i == n)
          add(f[i][j][k][0], f[i - 1][j][0][k] + f[i - 1][j][1][k]);
        else
          add(f[i][j][k][1], f[i - 1][j][0][k] + f[i - 1][j][1][k]);
        add(f[i][j][k][0], f[i - 1][j - 1][1][k]);
        if (j && i != n) add(f[i][j][k][0], f[i - 1][j - 1][0][k]);
        if (j && i != n) add(f[i][j][k][1], f[i - 1][j - 1][1][k]);
      }
  for (int i = (0); i <= (n); ++i)
    for (int j = (0); j <= (1); ++j)
      for (int k = (0); k <= (1); ++k) add(g[i], f[n][i][j][k]);
  for (int i = (0); i <= (n); ++i) g[i] = g[i] * jc[n - i] % mod;
  for (int i = (n); i >= (0); --i)
    for (int j = (i + 1); j <= (n); ++j)
      g[i] =
          (g[i] - g[j] * jc[j] % mod * inv[i] % mod * inv[j - i] % mod) % mod;
  printf("%d\n", (g[m] + mod) % mod);
  return 0;
}
