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 long long int MOD = 1000000007;
const long long BIG = 1446803456761533460;
const int Big = 336860180;
stringstream sss;
const long long int maxn = 1010;
long long int n, k;
long long int dp[maxn][maxn][4];
long long int first[maxn], ent[maxn][maxn], fact[maxn];
void MAIN() {
  fact[0] = 1;
  for (long long int i = (0); i < (maxn - 1); ++i)
    fact[i + 1] = fact[i] * (i + 1) % MOD;
  fill(ent[0], ent[0] + maxn, 1);
  for (long long int i = (0); i < (maxn - 1); ++i)
    for (long long int j = (0); j < (maxn - 1); ++j) {
      ent[i + 1][j + 1] = (ent[i][j] + ent[i + 1][j]) % MOD;
    }
  cin >> n >> k;
  dp[0][0][0] = 1;
  for (long long int i = (0); i < (n - 1); ++i) {
    for (long long int j = (0); j < (n + 1); ++j) {
      dp[i + 1][j][0] = ((dp[i + 1][j][0]) + (dp[i][j][0] + dp[i][j][1] +
                                              dp[i][j][2] + dp[i][j][3])) %
                        MOD;
      dp[i + 1][j + 1][1] =
          ((dp[i + 1][j + 1][1]) + (dp[i][j][0] + dp[i][j][1])) % MOD;
      dp[i + 1][j + 1][2] =
          ((dp[i + 1][j + 1][2]) + (dp[i][j][0] + dp[i][j][2])) % MOD;
      dp[i + 1][j + 2][3] = ((dp[i + 1][j + 2][3]) + (dp[i][j][0])) % MOD;
    }
  }
  for (long long int i = (n + 1) - 1; i >= (0); --i) {
    first[i] = 0;
    for (long long int j = (0); j < (4); ++j)
      first[i] = ((first[i]) + (dp[n - 1][i][j])) % MOD;
    first[i] = (1LL * (first[i]) * (fact[n - i])) % MOD;
    for (long long int j = (i + 1); j < (n + 1); ++j) {
      first[i] = ((first[i]) + (MOD - ent[i][j] * first[j] % MOD)) % MOD;
    }
  }
  cout << first[k] << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
7 4
    )";
  MAIN();
  return 0;
}
