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 GCC optimize(3, "Ofast", "inline")
#pragma GCC target("avx,avx2")
using namespace std;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const long long N = 1005, mod = 1e9 + 7;
long long f[N][N][2][2], g[N], ans[N], n, m, fac[N], inv[N];
long long C(long long n, long long m) {
  return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
signed main() {
  read(n);
  read(m);
  fac[0] = inv[0] = inv[1] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  for (long long i = 2; i <= n; i++)
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  for (long long i = 1; i <= n; i++) inv[i] = inv[i] * inv[i - 1] % mod;
  f[1][0][0][0] = f[1][1][0][1] = 1;
  for (long long i = 2; i <= n; i++)
    for (long long j = 0; j < i; j++) {
      f[i][j][0][0] =
          (f[i][j][0][0] + f[i - 1][j][0][0] + f[i - 1][j][1][0]) % mod;
      f[i][j][1][0] =
          (f[i][j][1][0] + f[i - 1][j][0][1] + f[i - 1][j][1][1]) % mod;
      f[i][j + 1][0][0] = f[i - 1][j][0][0];
      f[i][j + 1][0][1] = (f[i - 1][j][0][0] + f[i - 1][j][1][0]) % mod;
      f[i][j + 1][1][0] = f[i - 1][j][0][1];
      f[i][j + 1][1][1] = (f[i - 1][j][0][1] + f[i - 1][j][1][1]) % mod;
    }
  for (long long i = 0; i <= n; i++)
    g[i] = (f[n][i][0][0] + f[n][i][1][0]) * fac[n - i] % mod;
  for (long long i = 0; i <= n; i++)
    for (long long j = i, op = 1; j <= n; j++, op *= -1)
      ans[i] = (ans[i] + mod + op * C(j, i) * g[j] % mod) % mod;
  write(ans[m]);
}
