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 = 1e9 + 7;
long long n, k, dp[1111][1111][2][2], jc[2222], injc[2222], res[1111];
long long C(long long n, long long k) {
  if (n < k) return 0;
  return (jc[n] * injc[k] % mod) * injc[n - k] % mod;
}
long long binpow(long long a, long long t) {
  long long res = 1, p = a;
  for (long long i = t; i; i >>= 1) {
    if (i & 1) res = res * p % mod;
    p = p * p % mod;
  }
  return res;
}
int main() {
  scanf("%I64d%I64d", &n, &k);
  jc[0] = injc[0] = 1;
  for (long long i = 1; i <= 1000; i++) {
    jc[i] = jc[i - 1] * i % mod;
    injc[i] = injc[i - 1] * binpow(i, mod - 2) % mod;
  }
  dp[0][0][0][0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= i; j++) {
      for (long long k = 0; k <= 1; k++) {
        for (long long h = 0; h <= 1; h++) {
          if (!h)
            dp[i][j][k][h] = (dp[i - 1][j][0][k] + dp[i - 1][j][1][k]) % mod;
          if (j) {
            if (h)
              dp[i][j][k][h] = (dp[i][j][k][h] + dp[i - 1][j - 1][0][k] +
                                dp[i - 1][j - 1][1][k]) %
                               mod;
            if (i - 1 && !h)
              dp[i][j][k][h] = (dp[i][j][k][h] + dp[i - 1][j - 1][0][k]) % mod;
          }
        }
      }
    }
  }
  for (long long i = 0; i <= n; i++) {
    res[i] = (dp[n][i][0][0] + dp[n][i][1][0]) * jc[n - i] % mod;
  }
  for (long long i = n; i >= 0; i--) {
    for (long long j = i + 1; j <= n; j++) {
      res[i] = (res[i] - res[j] * C(j, i)) % mod;
    }
  }
  printf("%I64d\n", (res[k] + mod) % mod);
  return 0;
}
