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 int MAXN = 1020;
const long long mod = 1e9 + 7;
long long C[MAXN][MAXN], dp[MAXN][MAXN][2][2], ans[MAXN], fac[MAXN];
void add(long long &x, long long d) { x = (x + d) % mod; }
int main() {
  memset(C, 0, sizeof(C));
  fac[0] = fac[1] = 1;
  for (int i = 2; i < MAXN; ++i) fac[i] = fac[i - 1] * i % mod;
  for (int i = 0; i < MAXN; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
  }
  int n, t;
  scanf("%d%d", &n, &t);
  memset(dp, 0, sizeof(dp));
  dp[0][0][0][0] = 1;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= i; ++j)
      for (int k = 0; k < 2; ++k)
        for (int l = 0; l < 2; ++l)
          if (dp[i][j][k][l]) {
            long long &d = dp[i][j][k][l];
            if (!k && i) add(dp[i + 1][j + 1][l][0], d);
            if (i + 2 <= n) add(dp[i + 1][j + 1][l][1], d);
            add(dp[i + 1][j][l][0], d);
          }
  for (int i = 0; i <= n; ++i) {
    for (int k = 0; k < 2; ++k)
      for (int l = 0; l < 2; ++l) add(ans[i], dp[n][i][k][l]);
    ans[i] = ans[i] * fac[n - i] % mod;
  }
  for (int i = n; i >= t; --i)
    for (int j = i + 1; j <= n; ++j) ans[i] = (ans[i] - ans[j] * C[j][i]) % mod;
  printf("%lld\n", (ans[t] + mod) % mod);
  return 0;
}
