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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
const int P = 1e9 + 7;
int n, m, dp[1005][1005][2][2], fc[1005], fi[1005], F[1005], G[1005];
inline int ksm(int x, int q = P - 2) {
  int r = 1;
  for (; q; q >>= 1, x = 1ll * x * x % P)
    if (q & 1) r = 1ll * r * x % P;
  return r;
}
inline int C(int n, int m) {
  return n < 0 || m < 0 || n < m ? 0 : 1ll * fc[n] * fi[m] % P * fi[n - m] % P;
}
inline int pl(int a, int b) { return (a += b) >= P ? a -= P : a; }
int main() {
  read(n), read(m), fc[0] = 1;
  for (int i = 1; i <= n; i++) fc[i] = 1ll * fc[i - 1] * i % P;
  fi[n] = ksm(fc[n]);
  for (int i = n; i; i--) fi[i - 1] = 1ll * fi[i] * i % P;
  dp[1][0][0][0] = 1, dp[1][1][0][1] = 1;
  for (int i = 2; i < n; i++) {
    dp[i][0][0][0] = 1;
    for (int j = 1; j <= i; j++) {
      dp[i][j][0][0] = pl(pl(dp[i - 1][j - 1][0][0], dp[i - 1][j][0][0]),
                          dp[i - 1][j][1][0]);
      dp[i][j][1][0] = pl(pl(dp[i - 1][j - 1][0][1], dp[i - 1][j][0][1]),
                          dp[i - 1][j][1][1]);
      dp[i][j][0][1] = pl(dp[i - 1][j - 1][0][0], dp[i - 1][j - 1][1][0]);
      dp[i][j][1][1] = pl(dp[i - 1][j - 1][0][1], dp[i - 1][j - 1][1][1]);
    }
  }
  dp[n][0][0][0] = 1;
  for (int j = 1; j <= n; j++) {
    dp[n][j][0][0] =
        pl(pl(dp[n - 1][j - 1][0][0], dp[n - 1][j][0][0]), dp[n - 1][j][1][0]);
    dp[n][j][1][0] =
        pl(pl(dp[n - 1][j - 1][0][1], dp[n - 1][j][0][1]), dp[n - 1][j][1][1]);
  }
  for (int i = 0; i <= n; i++)
    F[i] = 1ll * (dp[n][i][0][0] + dp[n][i][1][0]) * fc[n - i] % P;
  int rs = 0;
  for (int i = m; i <= n; i++)
    rs = (rs + 1ll * ((i - m) & 1 ? P - 1ll : 1ll) * C(i, m) % P * F[i]) % P;
  return printf("%d\n", rs), 0;
}
