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;
long long n, K, f[1500][1500][2][2], num[1500];
long long ans[1500], fac[1500], C[1500][1500];
long long read() {
  long long x = 0, k = 1;
  char c;
  c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') k = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * k;
}
void Up(long long &x, long long y) { x = (x + y) % 1000000007; }
void pre() {
  fac[0] = 1;
  for (long long i = 1; i < 1500; i++) fac[i] = fac[i - 1] * i % 1000000007;
  for (long long i = 0; i < 1500; i++) C[i][0] = 1;
  for (long long i = 1; i < 1500; i++)
    for (long long j = 1; j < 1500; j++)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 1000000007;
}
signed main() {
  pre();
  n = read(), K = read();
  f[0][0][1][0] = 1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j <= i; j++)
      for (long long p = 0; p <= 1; p++)
        for (long long q = 0; q <= 1; q++) {
          Up(f[i + 1][j][q][0], f[i][j][p][q]);
          if (!p) Up(f[i + 1][j + 1][q][0], f[i][j][p][q]);
          if (i < n - 1) Up(f[i + 1][j + 1][q][1], f[i][j][p][q]);
        }
  }
  for (long long i = 0; i <= n; i++) {
    for (long long p = 0; p <= 1; p++)
      for (long long q = 0; q <= 1; q++) Up(num[i], f[n][i][p][q]);
    num[i] = num[i] * fac[n - i] % 1000000007;
  }
  for (long long i = n; i >= K; i--) {
    long long t = num[i];
    for (long long j = n; j > i; j--)
      t = (t - (ans[j] * C[j][i]) % 1000000007 + 1000000007) % 1000000007;
    ans[i] = t;
  }
  printf("%lld\n", ans[K]);
  return 0;
}
