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 MOD = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int n, K, C[1003][1003], fac[1003];
int dp[1003][1003][2][2], ans[1003];
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++)
      add(C[i][j], C[i - 1][j]), add(C[i][j], C[i - 1][j - 1]);
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % MOD;
  dp[0][0][1][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= i - 1; j++)
      for (int k = 0; k < 2; k++)
        for (int l = 0; l < 2; l++)
          if (dp[i - 1][j][k][l] > 0) {
            add(dp[i][j][l][0], dp[i - 1][j][k][l]);
            if (!k) add(dp[i][j + 1][l][0], dp[i - 1][j][k][l]);
            if (i < n) add(dp[i][j + 1][l][1], dp[i - 1][j][k][l]);
          }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++) add(ans[i], dp[n][i][j][k]);
    ans[i] = 1ll * ans[i] * fac[n - i] % MOD;
  }
  int cur = -1;
  for (int i = K + 1; i <= n; i++) {
    add(ans[K], (1ll * cur * C[i][K] % MOD * ans[i] % MOD + MOD) % MOD);
    cur *= -1;
  }
  printf("%d", ans[K]);
}
