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 NN = 1010;
const long long Mod = 1000000007;
long long N, K;
long long dp[NN][NN][2][2][2];
long long sum[NN];
long long fac[NN];
long long a[NN];
long long C[NN][NN];
void DP() {
  memset(dp, 0, sizeof(dp));
  dp[1][0][0][0][0] = 1;
  if (N > 1) dp[1][1][0][0][1] = 1;
  for (int i = 1; i < N; i++) {
    for (int j = 0; j <= i; j++) {
      for (int a = 0; a < 2; a++) {
        for (int b = 0; b < 2; b++) {
          for (int c = 0; c < 2; c++) {
            if (dp[i][j][a][b][c] == 0) continue;
            int tmp = dp[i][j][a][b][c];
            dp[i + 1][j][b][c][0] = (dp[i + 1][j][b][c][0] + tmp) % Mod;
            if (b == 0)
              dp[i + 1][j + 1][b][c][0] =
                  (dp[i + 1][j + 1][b][c][0] + tmp) % Mod;
            if (i + 2 <= N)
              dp[i + 1][j + 1][b][c][1] =
                  (dp[i + 1][j + 1][b][c][1] + tmp) % Mod;
          }
        }
      }
    }
  }
  for (int i = 0; i <= N; i++) {
    sum[i] = 0;
    sum[i] = dp[N][i][0][0][0] + dp[N][i][0][0][1] + dp[N][i][0][1][0] +
             dp[N][i][0][1][1] + dp[N][i][1][0][0] + dp[N][i][1][0][1] +
             dp[N][i][1][1][0] + dp[N][i][1][1][1];
    sum[i] %= Mod;
    sum[i] = sum[i] * fac[N - i] % Mod;
  }
}
int main() {
  fac[1] = fac[0] = 1;
  for (long long i = 2; i <= 1000; i++) fac[i] = fac[i - 1] * i % Mod;
  C[0][0] = C[1][0] = C[1][1] = 1;
  for (int i = 2; i <= 1000; i++) {
    for (int j = 0; j <= i; j++) {
      C[i][j] = C[i - 1][j - 1];
      if (i - 1 >= j) C[i][j] = (C[i][j] + C[i - 1][j]) % Mod;
    }
  }
  while (scanf("%d %d", &N, &K) == 2) {
    DP();
    a[N] = sum[N];
    for (int i = N - 1; i >= K; i--) {
      long long tmp = sum[i];
      for (int j = i + 1; j <= N; j++) {
        long long mid = C[j][i] * a[j];
        mid %= Mod;
        tmp -= mid;
        if (tmp < 0) tmp += Mod;
      }
      a[i] = tmp;
    }
    printf("%d\n", (int)a[K]);
  }
  return 0;
}
