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;
int Read() {
  char c;
  while (c = getchar(), (c != '-') && (c < '0' || c > '9'))
    ;
  bool neg = (c == '-');
  int ret = (neg ? 0 : c - 48);
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + c - 48;
  return neg ? -ret : ret;
}
const int MAXN = 1005, MOD = 7 + 1e9;
int N, K, f[MAXN], g[MAXN], h[MAXN], fac[MAXN], inv[MAXN];
int Pow(int x, int y) {
  if (y == 1) return x;
  int tmp = Pow(x, y >> 1);
  tmp = (long long)tmp * tmp % MOD;
  return y & 1 ? (long long)tmp * x % MOD : tmp;
}
int main() {
  scanf("%d%d", &N, &K), fac[0] = inv[0] = 1;
  for (int i = 1; i <= N; i++)
    fac[i] = (long long)fac[i - 1] * i % MOD, inv[i] = Pow(fac[i], MOD - 2);
  for (int i = 0; i <= N / 2; i++)
    g[i] = (long long)fac[N - i] * inv[i] % MOD * inv[N - i - i] % MOD;
  for (int i = 0; i <= N / 2 + 1; i++)
    for (int j = 0; j <= N / 2 + 1; j++)
      h[i + j] = ((long long)g[i] * g[j] + h[i + j]) % MOD;
  for (int i = 0; i <= N; i++) h[i] = (long long)h[i] * fac[N - i] % MOD;
  for (int i = 0; i <= N; i++)
    for (int j = 0; j <= i; j++)
      f[j] = ((long long)h[i] * fac[i] % MOD * inv[j] % MOD * inv[i - j] % MOD *
                  (1 - ((i - j) & 1) * 2) % MOD +
              MOD + f[j]) %
             MOD;
  printf("%d\n", f[K]);
  return 0;
}
