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;
template <class T>
inline void checkmin(T &a, const T &b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, const T &b) {
  if (b > a) a = b;
}
const int maxn = 1e3 + 10;
const int inf = 1e9 + 10;
const long long mod = 1e9 + 7;
int dp[maxn][maxn][4], g[maxn];
int n, k;
long long fac[maxn], c[maxn][maxn];
int main() {
  scanf("%d%d", &n, &k);
  dp[0][0][2] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= n; j++)
      for (int mask = 0; mask < 4; mask++) {
        if (dp[i][j][mask] == 0) continue;
        int mask2 = (mask & 1) << 1;
        dp[i + 1][j][mask2] += dp[i][j][mask];
        dp[i + 1][j][mask2] %= mod;
        if (!((mask >> 1) & 1))
          dp[i + 1][j + 1][mask2] += dp[i][j][mask],
              dp[i + 1][j + 1][mask2] %= mod;
        if (i == n - 1) continue;
        mask2 |= 1;
        dp[i + 1][j + 1][mask2] += dp[i][j][mask];
        dp[i + 1][j + 1][mask2] %= mod;
      }
  fac[0] = 1;
  for (long long i = 1; i < maxn; i++) fac[i] = fac[i - 1] * i % mod;
  for (int i = k; i <= n; i++) {
    long long now = 0;
    for (int mask = 0; mask < 4; mask++) now += dp[n][i][mask], now %= mod;
    g[i] = now * fac[n - i] % mod;
  }
  c[0][0] = 1;
  for (int i = 1; i < maxn; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
  }
  long long ans = 0;
  for (int i = k; i <= n; i++) {
    if ((i - k) & 1)
      ans -= c[i][k] * g[i] % mod, ans += mod;
    else
      ans += c[i][k] * g[i] % mod;
    ans %= mod;
  }
  printf("%d\n", ans);
  return 0;
}
