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 Mod = 1e9 + 7;
inline int Add(int x, int y) {
  x += y;
  return x >= Mod ? x - Mod : x;
}
inline void add(int &x, int y) { x = Add(x, y); }
inline int Sub(int x, int y) {
  x -= y;
  return x < 0 ? x + Mod : x;
}
inline void sub(int &x, int y) { x = Sub(x, y); }
inline int Mul(int x, int y) { return 1LL * x * y % Mod; }
inline void mul(int &x, int y) { x = Mul(x, y); }
int Pow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) mul(res, a);
    mul(a, a);
    b >>= 1;
  }
  return res;
}
const int N = 1010;
int n, k, f[N][N][2][2], fac[N], inv[N];
inline int C(int x, int y) {
  if (y > x) return 0;
  return Mul(fac[x], Mul(inv[x - y], inv[y]));
}
int main(void) {
  scanf("%d%d", &n, &k);
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = Mul(fac[i - 1], i);
  inv[n] = Pow(fac[n], Mod - 2);
  for (int i = n; i; --i) inv[i - 1] = Mul(inv[i], i);
  f[0][0][1][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= i; ++j) {
      add(f[i + 1][j][0][0], Add(f[i][j][0][0], f[i][j][1][0]));
      add(f[i + 1][j][1][0], Add(f[i][j][0][1], f[i][j][1][1]));
      add(f[i + 1][j + 1][0][0], f[i][j][0][0]);
      add(f[i + 1][j + 1][1][0], f[i][j][0][1]);
      add(f[i + 1][j + 1][0][1], Add(f[i][j][0][0], f[i][j][1][0]));
      add(f[i + 1][j + 1][1][1], Add(f[i][j][0][1], f[i][j][1][1]));
    }
  }
  int ans = 0;
  for (int j = k; j <= n; ++j) {
    int tmp = Add(f[n][j][0][0], f[n][j][1][0]);
    mul(tmp, Mul(fac[n - j], C(j, k)));
    if ((j - k) & 1)
      sub(ans, tmp);
    else
      add(ans, tmp);
  }
  printf("%d\n", ans);
  return 0;
}
