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 comment(linker, "/STACK:100000000,100000000")
using namespace std;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-9;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
long long f[1005][1005][1 << 2LL];
long long g[1005];
long long fact[1005];
long long c[1005][1005];
inline void add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main(void) {
  long long n, k;
  cin >> n >> k;
  f[0][0][2] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 0; j <= i; j++)
      for (long long m = 0; m < 4; m++) {
        if (m & 1) add(f[i][j + 1][m / 2 + 2], f[i - 1][j][m]);
        if (i < n) add(f[i][j + 1][m / 2], f[i - 1][j][m]);
        add(f[i][j][m / 2 + 2], f[i - 1][j][m]);
      }
  for (long long i = 0; i <= 1000; i++) {
    c[i][0] = c[i][i] = 1;
    for (long long j = 1; j < i; j++)
      if (i) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
  }
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) fact[i] = (fact[i - 1] * i) % mod;
  for (long long j = 0; j <= n; j++)
    for (long long m = 0; m < 4; m++) {
      add(g[j], (f[n][j][m] * fact[n - j]) % mod);
    }
  for (long long i = n - 1; i >= 0; i--) {
    for (long long j = i + 1; j <= n; j++) {
      long long sub = (g[j] * c[j][i]) % mod;
      g[i] = (g[i] - sub + mod) % mod;
    }
  }
  cout << g[k] << endl;
  return 0;
}
