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 N = 1005, mod = 1000000007;
inline void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int dp[N][N][2][2], f[N] = {1}, s[N], n, k, ans;
int qp(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return ans;
}
int c(int n, int m) {
  return 1ll * f[n] * qp(f[m], mod - 2) % mod * qp(f[n - m], mod - 2) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) f[i] = 1ll * f[i - 1] * i % mod;
  dp[0][0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      for (int x = 0; x < 2; x++) {
        for (int y = 0; y < 2; y++) {
          add(dp[i + 1][j][y][0], dp[i][j][x][y]);
          if (x == 0 && i != 0) add(dp[i + 1][j + 1][y][0], dp[i][j][x][y]);
          if (i != n - 1) add(dp[i + 1][j + 1][y][1], dp[i][j][x][y]);
        }
      }
    }
  }
  for (int j = 0; j <= n; j++) {
    for (int x = 0; x < 2; x++)
      for (int y = 0; y < 2; y++) add(s[j], dp[n][j][x][y]);
    s[j] = 1ll * s[j] * f[n - j] % mod;
  }
  ans = s[k];
  for (int i = k + 1; i <= n; i++) {
    if ((i - k) % 2)
      ans = (ans + mod - 1ll * s[i] * c(i, k) % mod) % mod;
    else
      ans = (ans + 1ll * s[i] * c(i, k) % mod) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
