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;
using ll = long long;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
const int mod = 1000000007;
ll dp[maxn][maxn][3];
ll f[maxn][maxn][2];
ll sum[maxn][maxn];
ll c1(int i, int j, int k) {
  if (j < 0) return 0;
  return dp[i][j][k];
}
ll c2(int i, int j, int k) {
  if (j < 0) return 0;
  return f[i][j][k];
}
ll c3(int i, int j) {
  if (j < 0) return 0;
  return sum[i][j];
}
int n, k;
int main() {
  cin >> n >> k;
  dp[2][2][2] = 1;
  f[2][0][0] = 1;
  sum[2][0] = sum[2][2] = sum[1][0] = 1;
  for (int i = 3; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      (dp[i][j][0] = c1(i - 1, j - 1, 0) + c1(i - 1, j, 1) + c1(i - 1, j, 2) +
                     c2(i - 1, j - 1, 1)) %= mod;
      (dp[i][j][1] = c1(i - 1, j, 0) + c1(i - 1, j - 1, 1) + c1(i - 1, j, 2) +
                     c2(i - 1, j - 1, 1)) %= mod;
      dp[i][j][2] = c3(i - 2, j - 2);
      f[i][j][0] = c3(i - 1, j);
      (f[i][j][1] =
           c2(i - 1, j + 1, 0) * (j + 1) +
           c2(i - 1, j, 0) * max(i - 1 - j - 1, 0) +
           c2(i - 1, j + 1, 1) * (j + 1) +
           c2(i - 1, j, 1) * max(i - 1 - j - 2, 0) + c1(i - 1, j + 1, 0) * j +
           c1(i - 1, j, 0) * max(i - 1 - j - 1, 0) + c1(i - 1, j + 1, 1) * j +
           c1(i - 1, j, 1) * max(i - 1 - j - 1, 0) +
           c1(i - 1, j + 1, 2) * max(j - 1, 0) +
           c1(i - 1, j, 2) * max(i - 1 - j, 0)) %= mod;
      sum[i][j] =
          (dp[i][j][0] + dp[i][j][1] + dp[i][j][2] + f[i][j][0] + f[i][j][1]) %
          mod;
    }
  }
  cout << sum[n][k] << "\n";
  return 0;
}
