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 int N = 1e3 + 5;
const int mod = 1e9 + 7;
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int mul(int a, int b) { return 1ll * a * b % mod; }
int inv(int a, int p) { return a == 1 ? 1 : p - 1ll * inv(p % a, a) * p / a; }
int Pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
int f[N][N][4];
int C[N][N];
int h[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int k;
  cin >> k;
  for (int i = 0; i <= n; ++i) C[i][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= i; ++j)
      add(C[i][j], C[i - 1][j]), add(C[i][j], C[i - 1][j - 1]);
  f[1][0][0] = 1;
  f[1][1][2] = (n > 1);
  for (int i = 2; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      for (int k = 0; k < 4; ++k) {
        int t0 = (k & 1);
        int t1 = (k & 2);
        if (t1) {
          if (!j) continue;
          add(f[i][j][k], f[i - 1][j - 1][t0 << 1]);
          add(f[i][j][k], f[i - 1][j - 1][t0 << 1 | 1]);
        } else {
          add(f[i][j][k], f[i - 1][j][t0 << 1]);
          add(f[i][j][k], f[i - 1][j][t0 << 1 | 1]);
          add(f[i][j][k], f[i - 1][j - 1][t0 << 1]);
        }
      }
  for (int i = 0; i <= n; ++i) {
    add(h[i], f[n][i][0]);
    add(h[i], f[n][i][1]);
    for (int j = 1; j <= n - i; ++j) h[i] = mul(h[i], j);
  }
  for (int i = n; i >= 0; --i)
    for (int j = n; j > i; --j) sub(h[i], mul(h[j], C[j][i]));
  cout << h[k] << endl;
}
