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 maxn = 1010;
const int Mod = 1e9 + 7;
int dp[maxn][maxn][2][2], fac[maxn], rev[maxn];
int qpow(int a, int x) {
  int res = 1;
  while (x) {
    if (x & 1) res = 1LL * res * a % Mod;
    x >>= 1;
    a = 1LL * a * a % Mod;
  }
  return res;
}
int C(int N, int M) { return 1LL * fac[N] * rev[M] % Mod * rev[N - M] % Mod; }
int main() {
  int N, M;
  scanf("%d%d", &N, &M);
  fac[0] = 1;
  rev[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = 1LL * fac[i - 1] * i % Mod;
  rev[N] = qpow(fac[N], Mod - 2);
  for (int i = N - 1; i >= 1; i--) rev[i] = 1LL * rev[i + 1] * (i + 1) % Mod;
  dp[0][0][0][0] = 1;
  for (int i = 0; i <= N - 1; i++) {
    for (int j = 0; j <= i; j++) {
      for (int p = 0; p <= 1; p++) {
        for (int q = 0; q <= 1; q++) {
          if (!dp[i][j][p][q]) continue;
          if (p == 0 && i) (dp[i + 1][j + 1][q][0] += dp[i][j][p][q]) %= Mod;
          (dp[i + 1][j + 1][q][1] += dp[i][j][p][q]) %= Mod;
          (dp[i + 1][j][q][0] += dp[i][j][p][q]) %= Mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = M; i <= N; i++) {
    int tmp = 1LL * (dp[N][i][1][0] + dp[N][i][0][0]) % Mod * C(i, M) % Mod *
              fac[N - i] % Mod;
    if ((i - M) % 2 == 0) {
      ans += tmp;
      if (ans >= Mod) ans -= Mod;
    } else {
      ans -= tmp;
      if (ans < 0) ans += Mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}
