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 MAX = 1e5 + 10;
const int MAXN = 1e4 + 10;
const long long MOD = 1e9 + 7;
const int inf = 1e9;
const double pi = acos(-1.0);
const double eps = 1e-6;
int dx[] = {0, -1, 0, 1};
int dy[] = {1, 0, -1, 0};
long long dp[1010][1010][2][2];
long long cnt[1010];
long long F[1010], invF[1010];
int n, k;
long long Pow(long long x, long long n) {
  if (n == 1) return x;
  long long tmp = Pow(x, n / 2);
  return ((tmp * tmp) % MOD * (n % 2 ? x : 1LL)) % MOD;
}
long long Comb(long long n, long long k) {
  long long res = 1LL;
  (res *= F[n]) %= MOD;
  (res *= invF[n - k]) %= MOD;
  (res *= invF[k]) %= MOD;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  F[0] = 1LL;
  invF[0] = 1LL;
  for (int i = 1; i <= n; i++) {
    F[i] = (F[i - 1] * (long long)i) % MOD;
    invF[i] = Pow(F[i], MOD - 2);
  }
  dp[0][0][1][0] = 1LL;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < i; j++)
      for (int x = 0; x < 2; x++)
        for (int y = 0; y < 2; y++) {
          if (y == 1) (dp[i][j + 1][1][x] += dp[i - 1][j][x][y]) %= MOD;
          if (i < n) (dp[i][j + 1][0][x] += dp[i - 1][j][x][y]) %= MOD;
          (dp[i][j][1][x] += dp[i - 1][j][x][y]) %= MOD;
        }
  for (int j = 0; j <= n; j++) {
    for (int x = 0; x < 2; x++)
      for (int y = 0; y < 2; y++) (cnt[j] += dp[n][j][x][y]) %= MOD;
    for (int x = 2; x <= n - j; x++) (cnt[j] *= (long long)x) %= MOD;
  }
  for (int i = n - 1; i >= 0; i--)
    for (int j = i + 1; j <= n; j++)
      cnt[i] = (cnt[i] - (cnt[j] * Comb(j, i)) % MOD + MOD) % MOD;
  cout << cnt[k];
  return 0;
}
