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>
const int MAX_SIZE = 1000;
const int MOD = 1000000000 + 7;
int n, m;
int C[MAX_SIZE + 10][MAX_SIZE + 10];
int f[MAX_SIZE + 10][MAX_SIZE + 10][4];
int fac[MAX_SIZE + 10], F[MAX_SIZE + 10];
inline void inc(int& a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int main() {
  scanf("%d%d", &n, &m);
  C[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
      if (C[i][j] >= MOD) C[i][j] -= MOD;
    }
  }
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (int)((long long)fac[i - 1] * i % MOD);
  f[0][0][1] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      for (int k = 0; k < 4; k++) {
        if (f[i - 1][j][k] == 0) continue;
        if (!(k & 1)) inc(f[i][j + 1][k >> 1], f[i - 1][j][k]);
        if (i < n) inc(f[i][j + 1][(k >> 1) | 2], f[i - 1][j][k]);
        inc(f[i][j][k >> 1], f[i - 1][j][k]);
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < 4; j++) inc(F[i], f[n][i][j]);
  }
  for (int i = n; i >= m; i--) {
    F[i] = (int)((long long)F[i] * fac[n - i] % MOD);
    for (int j = i + 1; j <= n; j++) {
      F[i] -= (int)((long long)F[j] * C[j][i] % MOD);
      if (F[i] < 0) F[i] += MOD;
    }
  }
  printf("%d\n", F[m]);
  return 0;
}
