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>
const int MAXN = 1010;
const int MOD = 1e9 + 7;
int DP[MAXN][MAXN][2][2], N, K, fac[MAXN], inv[MAXN];
inline long long C(int n, int r) {
  return (long long)fac[n] * inv[r] % MOD * inv[n - r] % MOD;
}
void add(int& x, int d) { (x += d) %= MOD; }
void init();
int main() {
  scanf("%d%d", &N, &K);
  init();
  DP[0][0][0][0] = 1;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= N; j++) {
      add(DP[i][j][0][0], DP[i - 1][j][0][0]);
      add(DP[i][j][0][0], DP[i - 1][j][0][1]);
      add(DP[i][j][0][0], DP[i - 1][j][1][0]);
      add(DP[i][j][0][0], DP[i - 1][j][1][1]);
      if (j >= 1 && i > 1) {
        add(DP[i][j][0][1], DP[i - 1][j - 1][0][0]),
            add(DP[i][j][0][1], DP[i - 1][j - 1][0][1]);
        add(DP[i][j][1][0], DP[i - 1][j - 1][0][0]),
            add(DP[i][j][1][0], DP[i - 1][j - 1][1][0]);
      }
      if (j >= 2 && i > 1) add(DP[i][j][1][1], DP[i - 1][j - 2][0][0]);
    }
  int ans = 0;
  for (int i = K; i <= N; i++)
    add(ans, (i - K & 1 ? -1LL : 1LL) * C(i, K) * fac[N - i] % MOD *
                 (((long long)DP[N][i][0][0] + DP[N][i][0][1] + DP[N][i][1][0] +
                   DP[N][i][1][1]) %
                  MOD) %
                 MOD);
  ans = (ans + MOD) % MOD;
  printf("%d\n", ans);
  return 0;
}
long long fast_pow(long long bs, int ex) {
  long long res = 1;
  for (; ex; (bs *= bs) %= MOD, ex >>= 1)
    if (ex & 1) (res *= bs) %= MOD;
  return res;
}
void init() {
  fac[0] = 1;
  for (long long i = 1; i <= N; i++) fac[i] = i * fac[i - 1] % MOD;
  inv[N] = fast_pow(fac[N], MOD - 2);
  for (long long i = N - 1; i >= 0; i--) inv[i] = (i + 1) * inv[i + 1] % MOD;
}
