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 rdtsc() {
  long long tmp;
  asm("rdtsc" : "=A"(tmp));
  return tmp;
}
const int MOD = (int)1e9 + 7;
inline void add(int &x, int toadd) {
  if ((x += toadd) >= MOD) x -= MOD;
}
inline void mult(int &x, int tomult) { x = (long long)x * tomult % MOD; }
const int maxn = 1001;
int C[maxn][maxn];
int fact[maxn];
int d[2][maxn][4];
void countFs(int n, int *fs) {
  for (int i = 0; i <= n; i++) memset(d[0][i], 0, sizeof(d[0][i]));
  d[0][0][2] = 1;
  for (int i = 0; i < n; i++) {
    int curit = (i & 1);
    for (int j = 0; j <= n; j++) memset(d[!curit][j], 0, sizeof(d[!curit][j]));
    for (int j = 0; j <= n; j++) {
      for (int mask = 0; mask < 4; mask++) {
        if (!d[curit][j][mask]) continue;
        if (!(mask >> 1))
          add(d[!curit][j + 1][(mask << 1) & 3], d[curit][j][mask]);
        add(d[!curit][j + 1][((mask << 1) ^ 1) & 3], d[curit][j][mask]);
        add(d[!curit][j][(mask << 1) & 3], d[curit][j][mask]);
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    fs[i] = d[n & 1][i][0];
    add(fs[i], d[n & 1][i][2]);
    mult(fs[i], fact[n - i]);
  }
}
int fs[maxn];
int values[maxn];
int main() {
  srand(rdtsc());
  memset(C, 0, sizeof(C));
  for (int n = 0; n < maxn; n++) {
    C[n][0] = C[n][n] = 1;
    for (int k = 1; k < n; k++)
      C[n][k] = C[n - 1][k - 1], add(C[n][k], C[n - 1][k]);
  }
  fact[0] = 1;
  for (int i = 1; i < maxn; i++) fact[i] = fact[i - 1], mult(fact[i], i);
  int n, k;
  while (scanf("%d%d", &n, &k) >= 1) {
    countFs(n, fs);
    int ans = 0;
    memset(values, 0, sizeof(values));
    for (int iter = k; iter <= n; iter++) {
      int coef = 0;
      for (int i = k; i < iter; i++) {
        int tmp = values[i];
        mult(tmp, C[iter][i]);
        add(coef, tmp);
      }
      coef = -coef;
      add(coef, MOD + (iter == k));
      values[iter] = coef;
      int tmp = fs[iter];
      mult(tmp, coef);
      add(ans, tmp);
    }
    printf("%d\n", ans);
  }
  return 0;
}
