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;
void add(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
int mul(int a, int b) { return 1LL * a * b % 1000000007; }
int n, k, dp[1010][1010][2][2][2];
int main() {
  scanf("%d%d", &n, &k);
  dp[1][0][0][0][1] = 1;
  for (int i = 1; i < n; i++)
    for (int j = 0; j < i + 1; j++)
      for (int x = 0; x < 2; x++)
        for (int y = 0; y < 2; y++)
          for (int z = 0; z < 2; z++) {
            int me = dp[i][j][x][y][z];
            add(dp[i + 1][j][0][0][1], me);
            if (z == 0) {
              add(dp[i + 1][j + 1 - x][0][1][0], me);
              add(dp[i + 1][j + 1 - y][1][0][0], me);
              add(dp[i + 1][j][0][0][0], mul(i - j + x + y - 2, me));
              add(dp[i + 1][j - 1][0][0][0], mul(j - x - y, me));
            } else {
              add(dp[i + 1][j + 2][1][1][0], me);
              add(dp[i + 1][j][0][0][0], mul(i - j - 1, me));
              add(dp[i + 1][j - 1][0][0][0], mul(j, me));
            }
          }
  int ans = 0;
  for (int x = 0; x < 2; x++)
    for (int y = 0; y < 2; y++)
      for (int z = 0; z < 2; z++) add(ans, dp[n][k][x][y][z]);
  printf("%d\n", ans);
  return 0;
}
