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 mod = 1000000007;
long long dp[1007][1007][4];
long long s[1007];
long long wyn[1007];
long long npok[1007][1007];
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  dp[1][1][1] = 1;
  dp[1][1][3] = 1;
  for (int i = 2; i <= N; i++) {
    for (int w = 0; w < 4; w++) dp[i][1][w] = 1;
    for (int w = 2; w <= N; w++) {
      dp[i][w][2] =
          (s[w - 1] - dp[i - 2][w - 1][1] + dp[i - 1][w - 1][2] + mod) % mod;
      dp[i][w][0] = (dp[i][w][2] + dp[i - 1][w - 1][3]) % mod;
      dp[i][w][3] = (s[w - 1] + dp[i - 1][w - 1][0]) % mod;
      dp[i][w][1] = (dp[i][w][3] + dp[i - 1][w - 1][1]) % mod;
      s[w - 1] += dp[i - 1][w - 1][0] + dp[i - 1][w - 1][1];
      s[w - 1] %= mod;
    }
  }
  npok[0][0] = 1;
  for (int i = 1; i <= N; i++) {
    npok[i][0] = 1;
    for (int w = 1; w <= N; w++)
      npok[i][w] = (npok[i - 1][w - 1] + npok[i - 1][w]) % mod;
  }
  int q = 1;
  long long silnia = 1;
  dp[N][0][0] = 1;
  for (int i = N; i >= K; i--) {
    wyn[i] = (dp[N][i][0] + s[i]) * silnia;
    wyn[i] %= mod;
    for (int w = i + 1; w <= N; w++) {
      wyn[i] -= (wyn[w] * npok[w][i]) % mod - mod;
      wyn[i] %= mod;
    }
    silnia *= q++;
    silnia %= mod;
  }
  printf("%I64d", wyn[K]);
  return 0;
}
