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 <class T>
inline void read(T& num) {
  num = 0;
  bool f = true;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = false;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    num = num * 10 + ch - '0';
    ch = getchar();
  }
  num = f ? num : -num;
}
int out[100];
template <class T>
inline void write(T x, char ch) {
  if (x == 0) {
    putchar('0');
    putchar(ch);
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int num = 0;
  while (x) {
    out[num++] = (x % 10);
    x = x / 10;
  }
  for (int i = (num - 1); i >= (0); i--) putchar(out[i] + '0');
  putchar(ch);
}
const int mod = 1000000007;
void add(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
long long ans[1010];
int C[1010][1010];
int dp[1010][1010][2][2];
int n, k;
int main() {
  cin >> n >> k;
  dp[0][0][1][0] = 1;
  for (int i = (1); i <= (n); i++)
    for (int j = (0); j <= (n - 1); j++)
      for (int now = (0); now <= (1); now++)
        for (int nxt = (0); nxt <= (1); nxt++) {
          if (!now) add(dp[i][j + 1][nxt][0], dp[i - 1][j][now][nxt]);
          if (i < n) add(dp[i][j + 1][nxt][1], dp[i - 1][j][now][nxt]);
          add(dp[i][j][nxt][0], dp[i - 1][j][now][nxt]);
        }
  for (int i = (0); i <= (n); i++)
    for (int j = (0); j <= (i); j++)
      C[i][j] = j ? (C[i - 1][j] + C[i - 1][j - 1]) % mod : 1;
  for (int i = (0); i <= (n); i++) {
    for (int j = (0); j <= (1); j++)
      for (int k = (0); k <= (1); k++) {
        ans[i] += dp[n][i][j][k];
        ans[i] %= mod;
      }
    for (int j = (2); j <= (n - i); j++) ans[i] = ans[i] * j % mod;
  }
  for (int i = (n); i >= (0); i--)
    for (int j = (i + 1); j <= (n); j++) {
      ans[i] -= ans[j] * C[j][i];
      ans[i] = (ans[i] % mod + mod) % mod;
    }
  cout << ans[k];
  return 0;
}
