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;
int len;
int num;
struct node {
  int d;
};
node dp[1100][1100][5];
void operator+=(node& a, node b) {
  a.d += b.d;
  if (a.d >= 1000000007) a.d -= 1000000007;
  return;
}
node operator+(node a, node b) {
  node ret;
  ret.d = a.d + b.d;
  if (ret.d >= 1000000007) ret.d -= 1000000007;
  return ret;
}
node operator*(node a, int b) {
  return (node){(int)((long long int)a.d * b % 1000000007)};
}
int main() {
  int f1;
  int f2;
  scanf("%d%d", &len, &num);
  dp[1][0][0].d = 1;
  for (f1 = 1; f1 < len; f1++)
    for (f2 = 0; f2 <= f1; f2++) {
      dp[f1 + 1][f2][0] += dp[f1][f2][0];
      dp[f1 + 1][f2][1] += dp[f1][f2][0] * (f1 - f2 - 1);
      dp[f1 + 1][f2 - 1][1] += dp[f1][f2][0] * f2;
      dp[f1 + 1][f2 + 2][4] += dp[f1][f2][0];
      dp[f1 + 1][f2][0] += dp[f1][f2][1];
      dp[f1 + 1][f2 + 1][2] += dp[f1][f2][1];
      dp[f1 + 1][f2 + 1][3] += dp[f1][f2][1];
      dp[f1 + 1][f2][1] += dp[f1][f2][1] * (f1 - f2 - 2);
      dp[f1 + 1][f2 - 1][1] += dp[f1][f2][1] * f2;
      dp[f1 + 1][f2][0] += dp[f1][f2][2];
      dp[f1 + 1][f2 + 1][2] += dp[f1][f2][2];
      dp[f1 + 1][f2][3] += dp[f1][f2][2];
      dp[f1 + 1][f2][1] += dp[f1][f2][2] * (f1 - f2 - 1);
      dp[f1 + 1][f2 - 1][1] += dp[f1][f2][2] * (f2 - 1);
      dp[f1 + 1][f2][0] += dp[f1][f2][3];
      dp[f1 + 1][f2][2] += dp[f1][f2][3];
      dp[f1 + 1][f2 + 1][3] += dp[f1][f2][3];
      dp[f1 + 1][f2][1] += dp[f1][f2][3] * (f1 - f2 - 1);
      dp[f1 + 1][f2 - 1][1] += dp[f1][f2][3] * (f2 - 1);
      dp[f1 + 1][f2][0] += dp[f1][f2][4];
      dp[f1 + 1][f2][2] += dp[f1][f2][4];
      dp[f1 + 1][f2][3] += dp[f1][f2][4];
      dp[f1 + 1][f2][1] += dp[f1][f2][4] * (f1 - f2);
      dp[f1 + 1][f2 - 1][1] += dp[f1][f2][4] * (f2 - 2);
    }
  printf("%d\n", (dp[len][num][0] + dp[len][num][1] + dp[len][num][2] +
                  dp[len][num][3] + dp[len][num][4])
                     .d);
  return 0;
}
