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 = (int)(1e9 + 7);
const int N = 1111;
long long f[N][N][4], result[N], C[N][N];
int n, kk;
inline void add(long long &a, const long long &b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int main() {
  scanf("%d %d", &n, &kk);
  for (int i = 0; i <= n; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
    }
  }
  f[0][0][1] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k < 4; k++)
        for (int l = 0; l < 4; l++) {
          if ((l >> 1) != (k & 1)) continue;
          if ((k >> 1)) {
            if (j) add(f[i][j][k], f[i - 1][j - 1][l]);
          } else {
            add(f[i][j][k], f[i - 1][j][l]);
            if ((l & 1) == 0 && j) add(f[i][j][k], f[i - 1][j - 1][l]);
          }
        }
    }
  for (int i = n; i >= kk; i--) {
    for (int j = 0; j < 2; j++) add(result[i], f[n][i][j]);
    for (int j = 1; j <= n - i; j++) result[i] = (result[i] * j) % MOD;
    for (int j = n; j > i; j--) {
      add(result[i], (MOD - result[j] * C[j][i] % MOD));
    }
  }
  printf("%I64d\n", result[kk]);
  return 0;
}
