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:102400000,102400000")
using namespace std;
int dp[(2000 + 20)][(2000 + 20)][5];
int f[(2000 + 20)], g[(2000 + 20)];
int C[(2000 + 20)][(2000 + 20)];
void add(int &u, int v) {
  u += v;
  if (u >= 1000000007) u -= 1000000007;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  f[0] = 1;
  for (int i = 1; i <= n; ++i) f[i] = (long long)f[i - 1] * i % 1000000007;
  C[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 1000000007;
  }
  dp[0][0][2] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < i; ++j) {
      for (int s = 0; s < 4; ++s) {
        add(dp[i][j][s / 2 + 2], dp[i - 1][j][s]);
        if (s & 1) add(dp[i][j + 1][s / 2 + 2], dp[i - 1][j][s]);
        if (i < n) add(dp[i][j + 1][s / 2], dp[i - 1][j][s]);
      }
    }
  }
  for (int i = n; i >= 0; --i) {
    for (int j = 0; j < 4; ++j) add(g[i], dp[n][i][j]);
    g[i] = (long long)g[i] * f[n - i] % 1000000007;
    for (int j = i + 1; j <= n; ++j) {
      g[i] -= (long long)g[j] * C[j][i] % 1000000007;
      if (g[i] < 0) g[i] += 1000000007;
    }
  }
  printf("%d\n", g[k]);
}
