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 N = 1010;
const int mo = 1e9 + 7;
int n, k, dp[N][N][4], ans[N], fac[N], fav[N];
int qpow(int a, int b) {
  int ans = 1;
  for (; b >= 1; b >>= 1, a = 1ll * a * a % mo)
    if (b & 1) ans = 1ll * ans * a % mo;
  return ans;
}
int C(int x, int y) {
  return x < y ? 0 : 1ll * fac[x] * fav[y] % mo * fav[x - y] % mo;
}
int main() {
  scanf("%d%d", &n, &k);
  if (n == k) {
    printf("%d\n", n & 1 ^ 1);
    return 0;
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mo;
  fav[n] = qpow(fac[n], mo - 2);
  for (int i = n; i >= 1; i--) fav[i - 1] = 1ll * fav[i] * i % mo;
  dp[0][0][1] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      for (int S = 0; S < 4; S++)
        if (dp[i - 1][j][S]) {
          (dp[i][j][S >> 1] += dp[i - 1][j][S]) %= mo;
          if (i != n) (dp[i][j + 1][(S >> 1) | 2] += dp[i - 1][j][S]) %= mo;
          if (S & 1 ^ 1) (dp[i][j + 1][S >> 1] += dp[i - 1][j][S]) %= mo;
        }
    }
  }
  for (int i = n; i >= k; i--) {
    ans[i] = (dp[n][i][0] + dp[n][i][1]) % mo * 1ll * fac[n - i] % mo;
    for (int j = i + 1; j <= n; j++) {
      ans[i] = (ans[i] - 1ll * ans[j] * C(j, i) % mo + mo) % mo;
    }
  }
  printf("%d\n", ans[k]);
  return 0;
}
