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;
long long n, m, ans;
long long f[2010], h[2010], c[2010][2010], dp[2010][2010][2][2];
long long read() {
  char ch = getchar();
  long long x = 0;
  while (ch > '9' || ch < '0') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
int main() {
  n = read();
  m = read();
  c[0][0] = 1, h[0] = 1;
  for (long long i = 1; i <= n; i++) {
    c[i][0] = 1, h[i] = h[i - 1] * i % 1000000007;
    for (long long j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007;
  }
  dp[0][0][1][0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 0; j < i; j++)
      for (long long now = 0; now < 2; now++)
        for (long long nxt = 0; nxt < 2; nxt++) {
          dp[i][j][nxt][0] =
              (dp[i][j][nxt][0] + dp[i - 1][j][now][nxt]) % 1000000007;
          if (!now)
            dp[i][j + 1][nxt][0] =
                (dp[i][j + 1][nxt][0] + dp[i - 1][j][now][nxt]) % 1000000007;
          if (i < n)
            dp[i][j + 1][nxt][1] =
                (dp[i][j + 1][nxt][1] + dp[i - 1][j][now][nxt]) % 1000000007;
        }
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; j < 2; j++)
      for (long long k = 0; k < 2; k++)
        f[i] = (f[i] + dp[n][i][j][k]) % 1000000007;
    f[i] = f[i] * h[n - i] % 1000000007;
  }
  for (long long i = m; i <= n; i++)
    ans = (ans + ((i - m) & 1 ? -1 : 1) * f[i] * c[i][m] + 1000000007) %
          1000000007;
  printf("%lld\n", ans);
  return 0;
}
