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;
long long dp[1100][1100][2][2], jc[1100], inv[1100], n, m, f[1100];
long long pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return ans;
}
long long C(long long n, long long m) {
  return jc[n] * inv[m] % MOD * inv[n - m] % MOD;
}
signed main() {
  scanf("%lld %lld", &n, &m);
  dp[0][0][1][0] = 1;
  for (long long i = 1; i <= n; i++)
    for (long long j = 0; j < i; j++)
      for (long long k = 0; k < 2; k++)
        for (long long p = 0; p < 2; p++) {
          dp[i][j][p][0] = (dp[i][j][p][0] + dp[i - 1][j][k][p]) % MOD;
          if (k == 0)
            dp[i][j + 1][p][0] =
                (dp[i][j + 1][p][0] + dp[i - 1][j][k][p]) % MOD;
          if (i < n)
            dp[i][j + 1][p][1] =
                (dp[i][j + 1][p][1] + dp[i - 1][j][k][p]) % MOD;
        }
  jc[0] = 1;
  for (long long i = 1; i <= n; i++) jc[i] = (jc[i - 1] * i) % MOD;
  inv[n] = pow(jc[n], MOD - 2);
  for (long long i = n - 1; i >= 0; i--) inv[i] = (inv[i + 1] * (i + 1)) % MOD;
  for (long long i = 0; i <= n; i++) {
    long long mid1 = 0;
    for (long long j = 0; j < 2; j++)
      for (long long k = 0; k < 2; k++) mid1 = (mid1 + dp[n][i][j][k]) % MOD;
    f[i] = mid1 % MOD * jc[n - i] % MOD;
  }
  long long ans = 0;
  for (long long i = m; i <= n; i++)
    ans = (ans + ((((i - m) & 1) ? -1 : 1) * C(i, m) % MOD * f[i] % MOD + MOD) %
                     MOD) %
          MOD;
  printf("%lld\n", ans);
  return 0;
}
