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 M = 1000000007;
int f[1000][1001][2][2];
int powmod(int a, int p) {
  if (p == 0) {
    return 1;
  } else if (p == 1) {
    return a;
  } else {
    long long t = powmod(a, p / 2);
    (t *= t) %= M;
    (t *= powmod(a, p % 2)) %= M;
    return t;
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= n; ++j) {
      for (int _1 = 0; _1 < 2; ++_1) {
        for (int _2 = 0; _2 < 2; ++_2) {
          f[i][j][_1][_2] = 0;
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    f[i][0][0][0] = 1;
    if (i == 0) {
      f[0][1][0][1] = 1;
      continue;
    }
    for (int j = 1; j <= i + 1; ++j) {
      for (int a = 0; a < 2; ++a) {
        f[i][j][a][0] += (f[i - 1][j][0][a] + f[i - 1][j][1][a]) % M;
        f[i][j][a][0] %= M;
        f[i][j][a][0] += f[i - 1][j - 1][0][a];
        f[i][j][a][0] %= M;
        f[i][j][a][1] += (f[i - 1][j - 1][0][a] + f[i - 1][j - 1][1][a]) % M;
        f[i][j][a][1] %= M;
      }
    }
  }
  vector<int> inv(n + 1);
  for (int i = 1; i <= n; ++i) {
    inv[i] = powmod(i, M - 2);
  }
  vector<long long> res(n + 1);
  for (int i = 0; i <= n; ++i) {
    res[i] = (f[n - 1][i][0][0] + f[n - 1][i][1][0]) % M;
  }
  for (int i = n - 1; i >= 0; --i) {
    for (int j = 1; j <= n - i; ++j) {
      (res[i] *= j) %= M;
    }
    long long cnk = 1;
    for (int j = i + 1; j <= n; ++j) {
      (cnk *= j) %= M;
      (cnk *= inv[j - i]) %= M;
      res[i] -= (res[j] * cnk) % M;
      (res[i] += M) %= M;
    }
  }
  cout << res[k] << endl;
  return 0;
}
