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 ___INT() {
  int ret;
  scanf("%d", &ret);
  return ret;
}
const int mod = 1000000007;
int dp[1005][1005][3][3];
int ncr[1005][1005];
int fact[1005];
int n, k;
void calcNcr() {
  ncr[0][0] = 1;
  for (int a = (1); a <= (1000); ++a)
    for (int b = (0); b <= (a); ++b) {
      if (b == 0)
        ncr[a][b] = 1;
      else
        ncr[a][b] = (ncr[a - 1][b - 1] + ncr[a - 1][b]) % mod;
    }
}
void calcFact() {
  fact[0] = 1;
  for (int i = 1; i <= n; ++i)
    fact[i] = ((long long)fact[i - 1] * (long long)i) % mod;
}
int F(int pos, int _k, int prev_prev, int prev) {
  if (_k < 0) return 0;
  if (pos == n) {
    if (_k == 0) return 1;
    return 0;
  }
  int& ret = dp[pos][_k][prev_prev][prev];
  if (ret > -1) return ret;
  ret = 0;
  if (pos != n - 1) ret = F(pos + 1, _k - 1, prev, 1);
  if (prev_prev != 1 && pos != 0)
    ret = (ret + F(pos + 1, _k - 1, prev, 0)) % mod;
  ret = (ret + F(pos + 1, _k, prev, 2)) % mod;
  return ret;
}
int calcSpecial(int _k) {
  long long A = F(0, _k, 2, 2);
  long long B = fact[n - _k];
  return (A * B) % mod;
}
int main() {
  n = ___INT();
  k = ___INT();
  memset(dp, (-1), sizeof(dp));
  calcNcr();
  calcFact();
  int special[1005];
  for (int _k = 0; _k <= n; ++_k) {
    special[_k] = calcSpecial(_k);
  }
  int normal[1005];
  for (int _k = n; _k >= 0; --_k) {
    normal[_k] = special[_k];
    for (int _k2 = _k + 1; _k2 <= n; ++_k2) {
      int invalids = ((long long)normal[_k2] * (long long)ncr[_k2][_k]) % mod;
      normal[_k] = (normal[_k] - invalids + mod) % mod;
    }
  }
  printf("%d\n", normal[k]);
  return 0;
}
