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;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
  return p1 == p2 &&
                 (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
             ? -1
             : *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 0;
  while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
  if (!x) return;
  recursive_print(x / 10);
  putc(x % 10 ^ 48);
}
template <typename T>
void print(T x) {
  if (!x) putc('0');
  if (x < 0) putc('-'), x = ~x + 1;
  recursive_print(x);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
}  // namespace fastio
const int MAXN = 1000;
const int MOD = 1e9 + 7;
int n, m, dp[MAXN + 5][MAXN + 5][2][2];
int fac[MAXN + 5], ifac[MAXN + 5];
void add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int binom(int x, int y) {
  return 1ll * fac[x] * ifac[y] % MOD * ifac[x - y] % MOD;
}
int main() {
  scanf("%d%d", &n, &m);
  dp[1][1][0][1] = dp[1][0][0][0] = 1;
  fac[0] = 1;
  ifac[0] = ifac[1] = 1;
  for (int i = 2; i <= n; i++)
    ifac[i] = 1ll * ifac[MOD % i] * (MOD - MOD / i) % MOD;
  for (int i = 1; i <= n; i++)
    fac[i] = 1ll * fac[i - 1] * i % MOD,
    ifac[i] = 1ll * ifac[i - 1] * ifac[i] % MOD;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j <= i; j++) {
      if (j) {
        add(dp[i][j][0][0], dp[i - 1][j - 1][0][0]);
        add(dp[i][j][1][0], dp[i - 1][j - 1][0][1]);
        add(dp[i][j][0][1], dp[i - 1][j - 1][0][0]);
        add(dp[i][j][0][1], dp[i - 1][j - 1][1][0]);
        add(dp[i][j][1][1], dp[i - 1][j - 1][0][1]);
        add(dp[i][j][1][1], dp[i - 1][j - 1][1][1]);
      }
      add(dp[i][j][0][0], dp[i - 1][j][0][0]);
      add(dp[i][j][0][0], dp[i - 1][j][1][0]);
      add(dp[i][j][1][0], dp[i - 1][j][0][1]);
      add(dp[i][j][1][0], dp[i - 1][j][1][1]);
    }
  int ans = 0;
  for (int i = m; i <= n; i++) {
    int ways = 1ll * (dp[n][i][0][0] + dp[n][i][1][0]) * fac[n - i] % MOD;
    if ((i - m) & 1)
      ans = (ans - 1ll * ways * binom(i, m) % MOD + MOD) % MOD;
    else
      ans = (ans + 1ll * ways * binom(i, m)) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}
