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;
inline long long qr() {
  long long sum = 0, f = 1;
  char y = getchar();
  while (!isdigit(y)) {
    if (y == '-') f = -1;
    y = getchar();
  }
  while (isdigit(y)) {
    sum = (sum << 3) + (sum << 1) + (y ^ 48);
    y = getchar();
  }
  return sum * f;
}
const long long N = 1005, mod = 1e9 + 7;
long long f[N][N][2][2], inv[N], mul[N];
long long n, m;
long long C(long long x, long long y) {
  return mul[x] * inv[x - y] % mod * inv[y] % mod;
}
long long qpow(long long a, long long b) {
  long long sum = 1;
  while (b) {
    if (b & 1) sum = sum * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return sum;
}
long long ans, F[N];
signed main() {
  n = qr(), m = qr();
  mul[0] = mul[1] = 1;
  inv[0] = 1;
  for (long long i = 2; i <= n; ++i) {
    mul[i] = mul[i - 1] * i % mod;
  }
  inv[n] = qpow(mul[n], mod - 2);
  for (long long i = n - 1; i > 0; --i) {
    inv[i] = (i + 1) * inv[i + 1] % mod;
  }
  f[1][0][0][0] = 1;
  f[1][1][0][1] = 1;
  for (long long i = 2; i <= n; ++i) {
    f[i][0][0][0] = 1;
    for (long long j = 1; j <= i; ++j) {
      f[i][j][0][0] = (f[i][j][0][0] + f[i - 1][j - 1][0][0]) % mod;
      f[i][j][1][0] += f[i - 1][j - 1][0][1] %= mod;
      (f[i][j][0][1] += f[i - 1][j - 1][0][0] + f[i - 1][j - 1][1][0]) %= mod;
      (f[i][j][1][1] += f[i - 1][j - 1][0][1] + f[i - 1][j - 1][1][1]) %= mod;
      (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 - 1][j][0][1] + f[i - 1][j][1][1]) %= mod;
    }
  }
  for (long long i = 0; i <= n; ++i) {
    F[i] = (f[n][i][0][0] + f[n][i][1][0]) % mod;
    F[i] = F[i] * mul[n - i] % mod;
  }
  long long tg = -1;
  for (long long i = m; i <= n; ++i) {
    tg *= -1;
    (ans += tg * C(i, m) * F[i]) %= mod;
  }
  ans = (ans % mod + mod) % mod;
  cout << ans;
  return 0;
}
