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 = 1005, P = 1000000007;
int n, m, dp[N][N][2][2], fac[N], ifac[N];
void upd(int &x, int y) { x = x + y >= P ? x + y - P : x + y; }
int pw(int a, int x) {
  int ans = 1;
  for (; x; x >>= 1, a = 1ll * a * a % P)
    if (x & 1) ans = 1ll * ans * a % P;
  return ans;
}
int C(int n, int m) { return 1ll * fac[n] * ifac[m] % P * ifac[n - m] % P; }
int main() {
  scanf("%d%d", &n, &m);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % P;
  ifac[n] = pw(fac[n], P - 2);
  for (int i = n; i; i--) ifac[i - 1] = 1ll * ifac[i] * i % P;
  dp[0][0][1][0] = 1;
  for (int i = 0; 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 (dp[i][j][x][y]) {
            upd(dp[i + 1][j][y][0], dp[i][j][x][y]);
            upd(dp[i + 1][j + 1][y][1], dp[i][j][x][y]);
            if (!x) upd(dp[i + 1][j + 1][y][0], dp[i][j][x][y]);
          }
  int ans = 0;
  for (int i = m, s = 1; i <= n; i++, s *= -1)
    ans = (ans + 1ll * s * (dp[n][i][0][0] + dp[n][i][1][0]) * C(i, m) % P *
                     fac[n - i]) %
          P;
  printf("%d\n", (ans + P) % P);
  return 0;
}
