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;
inline int read() {
  char x = '\0';
  int fh = 1, sum = 0;
  for (x = getchar(); x < '0' || x > '9'; x = getchar())
    if (x == '-') fh = -1;
  for (; x >= '0' && x <= '9'; x = getchar()) sum = sum * 10 + x - '0';
  return fh * sum;
}
const int N = 1009, mod = 1e9 + 7;
int n, m;
int f[N][N][2][2], F[N];
int jch[N], ijch[N];
inline int ksm(int a, int b) {
  int sum = 1;
  while (b) {
    if (b & 1) sum = 1LL * sum * a % mod;
    b >>= 1;
    a = 1LL * a * a % mod;
  }
  return sum;
}
inline void init() {
  jch[0] = 1;
  for (int i = 1; i <= n; i++) jch[i] = 1LL * jch[i - 1] * i % mod;
  ijch[n] = ksm(jch[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) ijch[i] = 1LL * ijch[i + 1] * (i + 1) % mod;
}
inline int C(int n, int m) {
  return 1LL * jch[n] * ijch[m] % mod * ijch[n - m] % mod;
}
int main() {
  n = read(), m = read();
  init();
  f[1][0][0][0] = 1, f[1][1][0][1] = 1;
  for (int i = 1; i <= n; i++) {
    f[i][0][0][0] = 1;
    for (int j = 1; j <= n; j++) {
      f[i][j][0][0] += f[i - 1][j - 1][0][0];
      f[i][j][1][0] += f[i - 1][j - 1][0][1];
      f[i][j][0][1] = (1LL * f[i][j][0][1] + f[i - 1][j - 1][0][0] +
                       f[i - 1][j - 1][1][0]) %
                      mod;
      f[i][j][1][1] = (1LL * f[i][j][1][1] + f[i - 1][j - 1][0][1] +
                       f[i - 1][j - 1][1][1]) %
                      mod;
      f[i][j][0][0] =
          (1LL * f[i][j][0][0] + f[i - 1][j][0][0] + f[i - 1][j][1][0]) % mod;
      f[i][j][1][0] =
          (1LL * f[i][j][1][0] + f[i - 1][j][0][1] + f[i - 1][j][1][1]) % mod;
    }
  }
  for (int i = 0; i <= n; i++)
    F[i] = 1LL * (f[n][i][1][0] + f[n][i][0][0]) * jch[n - i] % mod;
  int ans = 0;
  for (int i = m, fh = 1; i <= n; i++, fh *= -1)
    ans = (ans + 1LL * fh * C(i, m) * F[i] % mod + mod) % mod;
  printf("%d\n", ans);
  return 0;
}
