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 p = 1e9 + 7;
long long cs[1005][1005];
long long dp[1005][1005][2][2];
long long DP[1005];
int main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= i; j++)
      cs[i][j] = j ? (cs[i - 1][j] + cs[i - 1][j - 1]) % p : 1;
  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 kk = 0; kk < 2; kk++)
        for (long long l = 0; l < 2; l++) {
          if (!kk)
            dp[i][j + 1][l][0] = (dp[i][j + 1][l][0] + dp[i - 1][j][kk][l]) % p;
          if (i < n)
            dp[i][j + 1][l][1] = (dp[i][j + 1][l][1] + dp[i - 1][j][kk][l]) % p;
          dp[i][j][l][0] = (dp[i][j][l][0] + dp[i - 1][j][kk][l]) % p;
        }
  }
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; j < 2; j++)
      for (long long k = 0; k < 2; k++) DP[i] = (DP[i] + dp[n][i][j][k]) % p;
    for (long long j = 2; j <= n - i; j++) DP[i] = (DP[i] * j) % p;
  }
  for (long long i = n - 1; i >= 0; i--)
    for (long long j = i + 1; j <= n; j++) {
      DP[i] = (DP[i] - (cs[j][i] * DP[j]) % p + p) % p;
    }
  cout << DP[k] << endl;
}
