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 INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const long long mod = 1e9 + 7;
inline long long read() {
  long long ans = 0;
  char ch = getchar(), last = ' ';
  while (!isdigit(ch)) last = ch, ch = getchar();
  while (isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void MYFILE() {
  freopen("ha.in", "r", stdin);
  freopen("ha.out", "w", stdout);
}
int n, m;
long long dp[maxn][maxn][2][2], ans[maxn];
inline long long inc(long long a, long long b) {
  return a + b < mod ? a + b : a + b - mod;
}
long long fac[maxn], C[maxn][maxn];
inline void init() {
  fac[0] = 1;
  for (int i = 1; i < maxn; ++i) fac[i] = fac[i - 1] * i % mod;
  C[0][0] = 1;
  for (int i = 1; i < maxn; ++i) {
    C[i][0] = 1;
    for (int j = 1; j <= i; ++j) C[i][j] = inc(C[i - 1][j - 1], C[i - 1][j]);
  }
}
int main() {
  n = read(), m = read();
  init();
  dp[0][0][1][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < i; ++j)
      for (int k = 0; k < 2; ++k)
        for (int h = 0; h < 2; ++h) {
          dp[i][j][h][0] = inc(dp[i][j][h][0], dp[i - 1][j][k][h]);
          if (!k)
            dp[i][j + 1][h][0] = inc(dp[i][j + 1][h][0], dp[i - 1][j][k][h]);
          if (i < n)
            dp[i][j + 1][h][1] = inc(dp[i][j + 1][h][1], dp[i - 1][j][k][h]);
        }
  for (int i = 0; i <= n; ++i) {
    for (int k = 0; k < 2; ++k)
      for (int h = 0; h < 2; ++h) ans[i] = inc(ans[i], dp[n][i][k][h]);
    ans[i] = ans[i] * fac[n - i] % mod;
  }
  for (int i = m + 1; i <= n; ++i) {
    long long tp = C[i][m] * ans[i] % mod;
    ans[m] = inc(ans[m], ((i - m) & 1) ? mod - tp : tp);
  }
  write(ans[m]), puts("");
  return 0;
}
