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 dp[1001][1001][2][3];
int n, K;
long long sol(int i, int j, int a, int b) {
  if (j < 0) return 0;
  if (i == n) {
    if (j == K) return 1;
    return 0;
  }
  if (dp[i][j][a][b] != -1) return dp[i][j][a][b];
  long long rs = 0;
  if (b == 0 && a == 0) {
    rs += sol(i + 1, j, 0, 1);
    rs += sol(i + 1, j + 1, 1, 0);
    rs += sol(i + 1, j + 1, 0, 2);
    rs += j * sol(i + 1, j - 1, 0, 0);
    rs %= 1000000007;
    rs += (i - 2 - j) * sol(i + 1, j, 0, 0);
    rs %= 1000000007;
  } else if (b == 0 && a == 1) {
    rs += sol(i + 1, j, 0, 1);
    rs += sol(i + 1, j + 1, 1, 0);
    rs += sol(i + 1, j, 0, 2);
    rs += (j - 1) * sol(i + 1, j - 1, 0, 0);
    rs %= 1000000007;
    rs += (i - 1 - j) * sol(i + 1, j, 0, 0);
    rs %= 1000000007;
  } else if (b == 1 && a == 0) {
    rs += sol(i + 1, j, 0, 1);
    rs += sol(i + 1, j + 2, 1, 2);
    rs += j * sol(i + 1, j - 1, 0, 0);
    rs %= 1000000007;
    rs += (i - 1 - j) * sol(i + 1, j, 0, 0);
    rs %= 1000000007;
  } else if (b == 2 && a == 0) {
    rs += sol(i + 1, j, 0, 1);
    rs += sol(i + 1, j, 1, 0);
    rs += sol(i + 1, j + 1, 0, 2);
    rs += (j - 1) * sol(i + 1, j - 1, 0, 0);
    rs %= 1000000007;
    rs += (i - 1 - j) * sol(i + 1, j, 0, 0);
    rs %= 1000000007;
  } else if (b == 2 && a == 1) {
    rs += sol(i + 1, j, 0, 1);
    rs += sol(i + 1, j, 1, 0);
    rs += sol(i + 1, j, 0, 2);
    rs += (j - 2) * sol(i + 1, j - 1, 0, 0);
    rs %= 1000000007;
    rs += (i - j) * sol(i + 1, j, 0, 0);
    rs %= 1000000007;
  }
  return dp[i][j][a][b] = rs;
}
int main() {
  cin >> n >> K;
  if (n == 1) {
    if (K == 0) {
      cout << 1 << endl;
    } else
      cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < 1001; i++)
    for (int j = 0; j < 1001; j++)
      for (int a = 0; a < 2; a++)
        dp[i][j][a][0] = dp[i][j][a][1] = dp[i][j][a][2] = -1;
  cout << (sol(2, 0, 0, 1) + sol(2, 2, 1, 2)) % 1000000007 << endl;
}
