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 N = 1005, mod = 1e9 + 7;
long long n, k, dp[5][N][N], s;
int main() {
  scanf("%I64d%I64d", &n, &k);
  ;
  dp[0][1][0] = 1;
  dp[4][1][0] = 1;
  for (long long i = 2; i < n + 1; i++) {
    for (long long j = 0; j < n + 1; j++) {
      dp[0][i][j] = dp[4][i - 1][j];
      if (j > 1) dp[1][i][j] = dp[0][i - 1][j - 2];
      dp[2][i][j] = (dp[1][i - 1][j] * 2 + dp[2][i - 1][j]) % mod;
      if (j > 0)
        dp[2][i][j] += dp[2][i - 1][j - 1] + dp[3][i - 1][j - 1] * 2,
            dp[2][i][j] %= mod;
      dp[3][i][j] =
          (dp[0][i - 1][j] * max(i - 2 - j, s) + dp[0][i - 1][j + 1] * (j + 1) +
           dp[1][i - 1][j + 1] * max(j - 1, s) +
           dp[1][i - 1][j] * max(i - 1 - j, s));
      dp[3][i][j] %= mod;
      dp[3][i][j] += dp[2][i - 1][j] * max(i - 1 - j - 1, s) +
                     dp[2][i - 1][j + 1] * j + dp[3][i - 1][j + 1] * (j + 1) +
                     dp[3][i - 1][j] * max(i - j - 3, s);
      dp[3][i][j] %= mod;
      dp[4][i][j] =
          (dp[0][i][j] + dp[1][i][j] + dp[2][i][j] + dp[3][i][j]) % mod;
    }
  }
  cout << dp[4][n][k];
}
