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 long long MOD = 1000000007;
const int SIZEN = 1010;
long long dp[SIZEN][SIZEN][4];
long long fact[SIZEN] = {0};
long long C[SIZEN][SIZEN] = {0};
long long F[SIZEN] = {0}, G[SIZEN] = {0};
int N, K;
void add(long long &a, long long b) {
  a += b;
  a %= MOD;
  if (a < 0) a += MOD;
}
void work(void) {
  fact[0] = 1;
  for (int i = 1; i <= N; i++) fact[i] = fact[i - 1] * i % MOD;
  for (int i = 0; i <= N; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
    }
  }
  dp[0][0][0] = 1;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j <= i; j++) {
      for (int s = 0; s < 4; s++) {
        add(dp[i + 1][j][(s * 2) & 3], dp[i][j][s]);
        if (i > 0 && (s & 2) == 0) {
          add(dp[i + 1][j + 1][(s * 2) & 3], dp[i][j][s]);
        }
        if (i + 2 <= N) {
          add(dp[i + 1][j + 1][(s * 2 + 1) & 3], dp[i][j][s]);
        }
      }
    }
  }
  for (int i = 0; i <= N; i++) {
    for (int s = 0; s < 4; s++) {
      add(F[i], dp[N][i][s]);
    }
    F[i] = F[i] * fact[N - i] % MOD;
  }
  for (int i = N; i >= 0; i--) {
    G[i] = F[i];
    for (int j = N; j > i; j--) {
      add(G[i], -G[j] * C[j][i] % MOD);
    }
  }
  printf("%d\n", (int)G[K]);
}
int main() {
  scanf("%d%d", &N, &K);
  work();
  return 0;
}
