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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
const long long N = 2e3 + 100, OO = 1e15 + 7, T = 13, M = 1e9 + 7, P = 6151,
                SQ = 360, LG = 50;
long long dp[N][N][5], sum[N][N];
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, k;
  cin >> n >> k;
  dp[1][0][0] = 1;
  sum[1][0] = 1;
  sum[0][0] = 1;
  for (long long i = 2; i <= n; i++) {
    for (long long j = 0; j <= i; j++) {
      dp[i][j][0] = (((sum[i - 1][j] + dp[i - 1][j + 1][0] * (j + 1)) % M +
                      dp[i - 1][j + 1][1] * j) %
                         M +
                     dp[i - 1][j + 1][2] * j) %
                    M;
      if (j) dp[i][j][0] = (dp[i][j][0] + dp[i - 1][j + 1][3] * (j - 1)) % M;
      if (i - j - 3 > 0)
        dp[i][j][0] =
            (dp[i][j][0] + ((dp[i - 1][j][0] % M - sum[i - 2][j] % M + M) % M) *
                               (i - j - 3)) %
            M;
      if (i - j - 2 > 0)
        dp[i][j][0] =
            (dp[i][j][0] +
             ((dp[i - 1][j][1] + dp[i - 1][j][2] + sum[i - 2][j]) % M) *
                 (i - j - 2)) %
            M;
      if (i - j - 1 > 0)
        dp[i][j][0] = (dp[i][j][0] + dp[i - 1][j][3] * (i - j - 1)) % M;
      if (i >= 2 && j >= 2) dp[i][j][3] = sum[i - 2][j - 2];
      if (j)
        dp[i][j][1] = ((dp[i - 1][j - 1][0] - sum[i - 2][j - 1] + M) % M +
                       dp[i - 1][j - 1][1]) %
                      M;
      dp[i][j][1] = (dp[i][j][1] + dp[i - 1][j][2] + dp[i - 1][j][3]) % M;
      if (j)
        dp[i][j][2] = ((dp[i - 1][j - 1][0] - sum[i - 2][j - 1] + M) % M +
                       dp[i - 1][j - 1][2]) %
                      M;
      dp[i][j][2] = (dp[i][j][2] + dp[i - 1][j][1] + dp[i - 1][j][3]) % M;
      sum[i][j] = (dp[i][j][0] + dp[i][j][1] + dp[i][j][2] + dp[i][j][3]) % M;
    }
  }
  cout << sum[n][k] << endl;
  return 0;
}
