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 maxn = 1010;
const long long mo = 1000000007;
int n, m;
long long f[maxn], g[maxn], c[maxn][maxn], dp[maxn][maxn][2][2];
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    memset(g, 0, sizeof(g));
    memset(f, 0, sizeof(f));
    memset(c, 0, sizeof(c));
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i <= n; ++i) dp[i][0][0][0] = 1;
    f[0] = 1;
    for (int i = 1; i <= n; ++i) f[i] = f[i - 1] * i % mo;
    for (int i = 0; i <= n; ++i) c[i][0] = c[i][i] = 1;
    for (int i = 0; i <= n; ++i)
      for (int j = 1; j < n; ++j)
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mo;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= i; ++j) {
        if (i > 1)
          dp[i][j][0][0] = (dp[i - 1][j][0][0] + dp[i - 1][j][1][0] +
                            dp[i - 1][j - 1][0][0]) %
                           mo;
        dp[i][j][1][0] =
            (dp[i - 1][j][0][1] + dp[i - 1][j][1][1] + dp[i - 1][j - 1][0][1]) %
            mo;
        if (i < n) {
          dp[i][j][0][1] =
              (dp[i - 1][j - 1][0][0] + dp[i - 1][j - 1][1][0]) % mo;
          dp[i][j][1][1] =
              (dp[i - 1][j - 1][0][1] + dp[i - 1][j - 1][1][1]) % mo;
        }
      }
    for (int i = 0; i <= n; ++i)
      g[i] =
          (dp[n][i][0][0] + dp[n][i][0][1] + dp[n][i][1][0] + dp[n][i][1][1]) *
          f[n - i] % mo;
    long long ans = 0;
    for (int i = 0; i <= n - m; ++i)
      ans = (ans + c[m + i][m] * g[m + i] * (i & 1 ? -1 : 1) + mo) % mo;
    printf("%I64d\n", ans);
  }
  return 0;
}
