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 mod = 1e9 + 7;
const int N = 1005;
int n, m;
long long dp[N][N][2][2], fac[N], inv[N], ret[N], ans = 0;
long long c(int n, int m) { return fac[n] * inv[m] % mod * inv[n - m] % mod; }
int main() {
  scanf("%d%d", &n, &m);
  fac[0] = 1ll;
  inv[0] = inv[1] = 1ll;
  for (int i = 2; i <= n; i++) inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = inv[i] * inv[i - 1] % mod;
  }
  dp[1][0][0][0] = 1ll;
  dp[1][1][0][1] = 1ll;
  for (int i = 2; i < n; ++i) {
    dp[i][0][0][0] = 1;
    for (int j = 1; j <= i; ++j) {
      dp[i][j][0][0] = (dp[i][j][0][0] + dp[i - 1][j - 1][0][0] +
                        dp[i - 1][j][0][0] + dp[i - 1][j][1][0]) %
                       mod;
      dp[i][j][0][1] =
          (dp[i][j][0][1] + dp[i - 1][j - 1][0][0] + dp[i - 1][j - 1][1][0]) %
          mod;
      dp[i][j][1][0] = (dp[i][j][1][0] + dp[i - 1][j - 1][0][1] +
                        dp[i - 1][j][0][1] + dp[i - 1][j][1][1]) %
                       mod;
      dp[i][j][1][1] =
          (dp[i][j][1][1] + dp[i - 1][j - 1][0][1] + dp[i - 1][j - 1][1][1]) %
          mod;
    }
  }
  dp[n][0][0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    dp[n][i][0][0] = (dp[n][i][0][0] + dp[n - 1][i - 1][0][0] +
                      dp[n - 1][i][0][0] + dp[n - 1][i][1][0]) %
                     mod;
    dp[n][i][1][0] = (dp[n][i][1][0] + dp[n - 1][i - 1][0][1] +
                      dp[n - 1][i][0][1] + dp[n - 1][i][1][1]) %
                     mod;
  }
  for (int i = 0; i <= n; ++i) {
    ret[i] = (dp[n][i][0][0] + dp[n][i][1][0]) * fac[n - i] % mod;
  }
  for (int i = m; i <= n; ++i) {
    long long tot = ret[i] * c(i, m) % mod;
    ans = (ans + ((i - m) & 1 ? (mod - tot) : tot)) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
