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 MAX_N = 1005;
long long dp[MAX_N][MAX_N][2][2];
int n, k;
long long F[MAX_N], invF[MAX_N];
long long C[MAX_N][MAX_N];
long long cnt[MAX_N];
void add(long long &a, long long b) { (a += b) %= MOD; }
void init() {
  F[0] = 1;
  for (int i = 1; i <= n; i++) {
    F[i] = (1LL * F[i - 1] * i) % MOD;
  }
  for (int i = 0; i <= n; i++) {
    C[i][0] = C[i][i] = 1LL;
    for (int j = 1; j < i; j++) {
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  init();
  dp[0][0][1][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= n; j++) {
      for (int x = 0; x <= 1; x++) {
        for (int y = 0; y <= 1; y++) {
          if (dp[i][j][x][y] > 0) {
            long long cur = dp[i][j][x][y];
            add(dp[i + 1][j][y][0], cur);
            if (x == 0) {
              add(dp[i + 1][j + 1][y][0], cur);
            }
            if (i < n) {
              add(dp[i + 1][j + 1][y][1], cur);
            }
          }
        }
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    cnt[i] = (dp[n][i][0][0] + dp[n][i][1][0]) % MOD;
    (cnt[i] *= F[n - i]) %= MOD;
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i + 1; j <= n; j++) {
      cnt[i] = (cnt[i] - (C[j][i] * cnt[j]) % MOD + MOD) % MOD;
    }
  }
  printf("%d", cnt[k]);
  return 0;
}
