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>
#pragma GCC optimize("O2")
using namespace std;
const int N = 1e3 + 10;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long long inf = 8e18;
const int LOG = 22;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
long long Ans[N], dp[N][N][2][2], first[N], I[N];
inline long long E(long long r, long long n) {
  if (r > n || r < 0) return 0;
  return ((first[n] * I[r] % mod) * I[n - r] % mod);
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  first[0] = I[0] = 1;
  for (int i = 1; i < N; i++) {
    first[i] = first[i - 1] * i % mod;
    I[i] = pw(first[i], mod - 2, mod);
  }
  dp[0][0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      for (int w1 = 0; w1 < 2; w1++) {
        for (int w2 = 0; w2 < 2; w2++) {
          if (i == 0 && w1 == 1) continue;
          dp[i + 1][j + 1][w2][1] =
              (dp[i + 1][j + 1][w2][1] + dp[i][j][w1][w2]) % mod;
          if (i && w1 == 0) {
            dp[i + 1][j + 1][w2][0] =
                (dp[i + 1][j + 1][w2][0] + dp[i][j][w1][w2]) % mod;
          }
          dp[i + 1][j][w2][0] = (dp[i + 1][j][w2][0] + dp[i][j][w1][w2]) % mod;
        }
      }
    }
  }
  for (int j = 0; j <= n; j++) {
    for (int w1 = 0; w1 < 2; w1++) {
      Ans[j] = (Ans[j] + dp[n][j][w1][0]) % mod;
    }
    Ans[j] = (Ans[j] * first[n - j]) % mod;
  }
  for (int j = n - 1; j >= 0; j--) {
    for (int j2 = j + 1; j2 <= n; j2++) {
      Ans[j] = (Ans[j] - (Ans[j2] * E(j, j2) % mod) + mod) % mod;
    }
  }
  printf("%lld", Ans[k]);
  return 0;
}
