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 mod = 1e9 + 7;
const int N = 1020;
int f[2][N][2][2];
int g[N];
int fac[N], inv[N];
int n, m;
int C(int n, int m) {
  return (1LL * fac[n] * (1LL * inv[m] * inv[n - m] % mod) % mod);
}
int ksm(int a, int b = mod - 2) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = (1LL * ret * a % mod);
    a = (1LL * a * a % mod);
    b >>= 1;
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  f[1][0][0][0] = 1;
  f[1][1][0][1] = 1;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (1LL * fac[i - 1] * i % mod);
  inv[n] = ksm(fac[n]);
  for (int i = n - 1; i >= 0; i--) inv[i] = (1LL * inv[i + 1] * (i + 1) % mod);
  for (int cur = 2; cur <= n; cur++) {
    int i = cur & 1;
    for (int j = 0; j <= n; j++) {
      for (int l = 0; l < 2; l++)
        for (int k = 0; k < 2; k++) f[i][j][k][l] = 0;
      for (int k = 0; k < 2; k++)
        f[i][j][k][0] = (f[i][j][k][0] + f[i ^ 1][j - 1][0][k]) % mod;
      if (cur != n)
        for (int k = 0; k < 2; k++)
          for (int l = 0; l < 2; l++)
            f[i][j][k][1] = (f[i][j][k][1] + f[i ^ 1][j - 1][l][k]) % mod;
      for (int k = 0; k < 2; k++)
        for (int l = 0; l < 2; l++)
          f[i][j][k][0] = (f[i][j][k][0] + f[i ^ 1][j][l][k]) % mod;
    }
  }
  for (int j = m; j <= n; j++)
    for (int k = 0; k < 2; k++) g[j] = (g[j] + f[n & 1][j][k][0]) % mod;
  for (int j = m; j <= n; j++) g[j] = (1LL * g[j] * fac[n - j] % mod);
  int ans = 0;
  for (int i = 0; i + m <= n; i++) {
    int j = i + m;
    if (i & 1)
      ans = (ans + -(1LL * C(j, m) * g[j] % mod)) % mod;
    else
      ans = (ans + (1LL * C(j, m) * g[j] % mod)) % mod;
  }
  ans = (ans + mod) % mod;
  printf("%d\n", ans);
  return 0;
}
