Problem:
Score : 500 points

Problem Statement

Given are N integers A_1,\ldots,A_N.

We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.

Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Constraints

1 \leq K \leq N \leq 2\times 10^5

|A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 \ldots A_N

Output

Print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Sample Input 1

4 2
1 2 -3 -4

Sample Output 1

12

The possible products of the two chosen elements are 2, -3, -4, -6, -8, and 12, so the maximum product is 12.

Sample Input 2

4 3
-1 -2 -3 -4

Sample Output 2

1000000001

The possible products of the three chosen elements are -24, -12, -8, and -6, so the maximum product is -6.

We print this value modulo (10^9+7), that is, 1000000001.

Sample Input 3

2 1
-1 1000000000

Sample Output 3

1000000000

The possible products of the one chosen element are -1 and 1000000000, so the maximum product is 1000000000.

Sample Input 4

10 10
1000000000 100000000 10000000 1000000 100000 10000 1000 100 10 1

Sample Output 4

999983200

Be sure to print the product modulo (10^9+7).

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define BIT(n) (1LL << (n))
#define BITF(n, i) (((n) >> (i)) & 1)
#define REP(i, n) for (int i = 0; i < n; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define REPI(i, x) for (int i = 1; i <= x; i++)
#define FORI(i, m, n) for (int i = m; i <= n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define REPZ(i, x) for (int i = 0; i <= x; i++)
#define FORA(i, n) for (auto &&i : n)
#define POW(a, b) ((int)(pow(a, b) + .5))
#define MODULO(a, b) (((a) % (b)) < 0 ? (a) % (b) + (b) : (a) % (b))
#define DUMPOUT cerr
// vector
template <typename T> istream &operator>>(istream &is, vector<T> &vec) {
  for (T &x : vec) is >> x;
  return is;
}
// pair
template <typename T, typename U> ostream &operator<<(ostream &os, pair<T, U> &pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
// vector
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << "{";
  REP(i, (int)vec.size())
  os << vec[i] << (i + 1 == (int)vec.size() ? "" : ", ");
  os << "}";
  return os;
}
// map
template <typename T, typename U> ostream &operator<<(ostream &os, map<T, U> &map_var) {
  os << "{";
  FORA(itr, map_var) {
    os << *itr;
    itr++;
    if (itr != map_var.end()) os << ", ";
    itr--;
  }
  os << "}";
  return os;
}
// set
template <typename T> ostream &operator<<(ostream &os, set<T> &set_var) {
  os << "{";
  FORA(itr, set_var) {
    os << *itr;
    itr++;
    if (itr != set_var.end()) os << ", ";
    itr--;
  }
  os << "}";
  return os;
}
void dump_func() { DUMPOUT << endl; }
template <class Head, class... Tail> void dump_func(Head &&head, Tail &&... tail) {
  DUMPOUT << head;
  if (sizeof...(Tail) > 0) DUMPOUT << ", ";
  dump_func(std::move(tail)...);
}
#ifdef DEBUG_
#define DEB
#define DUMP(...)                                                             \
  DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                             \
          << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]" << endl \
          << "    ",                                                          \
      dump_func(__VA_ARGS__)
#define PRINTARR(x, y)                                     \
  cerr << #x << "=\n";                                     \
  for (auto itr = x; itr != y; itr++) cerr << *itr << " "; \
  cerr << endl;
#define PRINTARR2(x, i0, i1)                                       \
  cerr << #x << "=\n";                                             \
  for (int ii0 = 0; ii0 < i0; ii0++) {                             \
    for (int ii1 = 0; ii1 < i1; ii1++) cerr << x[ii0][ii1] << " "; \
    cerr << endl;                                                  \
  }
#else
#define DEB if (false)
#define DUMP(...)
#define PRINTARR(x, y)
#define PRINTARR2(x, i0, i1)
#endif
#define ALL(v) v.begin(), v.end()
#define fst first
#define snd second
#define mp make_pair
#define pb push_back
#define epb emplace_back
#define int long long
#define pint pair<int, int>
#define pdouble pair<double, double>
#define ld long double
using namespace std;
template <class T> bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> using vec = std::vector<T>;
template <class T> void print(const T &x) { cout << x << "\n"; }
const int MOD = 1000000007, INF0 = 1061109567, INF = INF0 * INF0;
const double EPS = 1e-7, PI = acos(-1.0);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
/*
[&] { return; }();下のスラッシュを2つ追加しただけ.上のスラッシュを2つにすれば外れる.
//*/
#define MAXN 200200
///！ここから
int binpow(int a, int b, int mo) {
  if (a >= mo) a %= mo;
  int ans = 1;
  do {
    if (b & 1) ans = 1ll * ans * a % mo;
    a = 1ll * a * a % mo;
  } while (b >>= 1);
  return ans;
}
int modInv(int p) {
  if (p >= MOD) p %= MOD;
  return binpow(p, MOD - 2, MOD);
}

signed main() {
  cin.tie(0), ios::sync_with_stdio(false);
  cout << fixed << setprecision(12);
  vec<pint> a;
  int N, K;
  cin >> N >> K;
  int zcnt = 0;
  REP(i, N) {
    int x;
    cin >> x;
    if (x > 0)
      a.pb(mp(x, 1));
    else if (x < 0)
      a.pb(mp(-x, -1));
    else
      zcnt++;
  }
  sort(ALL(a));
  reverse(ALL(a));
  DUMP(N, zcnt, K);
  if (zcnt > 0 && N - zcnt < K) {
    print(0);
    return 0;
  }
  int ans = 1;
  int cntM = 0;
  int pm0[2] = {-1, -1};  //+-
  int pm1[2] = {-1, -1};  //+-

  REP(i, K) {
    pm0[a[i].snd == -1] = i;
    cntM += a[i].snd == -1;
    ans *= a[i].fst;
    ans %= MOD;
  }
  for (int i = a.size() - 1; i >= K; i--) {
    pm1[a[i].snd == -1] = i;
  }
  if (cntM % 2 == 0) {
    print(ans);
  } else {
    if (pm0[0] != -1 && pm0[1] != -1 && pm1[0] != -1 && pm1[1] != -1) {
      int a0 = a[pm0[0]].fst;
      int a1 = a[pm0[1]].fst;
      int b0 = a[pm1[0]].fst;
      int b1 = a[pm1[1]].fst;
      if (b1 * a1 > b0 * a0) {
        ans *= b1;
        ans %= MOD;
        ans *= modInv(a0);
        ans %= MOD;
      } else {
        ans *= b0;
        ans %= MOD;
        ans *= modInv(a1);
        ans %= MOD;
      }
    } else if (pm0[0] != -1 && pm1[1] != -1) {
      int a0 = a[pm0[0]].fst;
      int b1 = a[pm1[1]].fst;
      ans *= b1;
      ans %= MOD;
      ans *= modInv(a0);
      ans %= MOD;
    } else if (pm0[1] != -1 && pm1[0] != -1) {
      int a1 = a[pm0[1]].fst;
      int b0 = a[pm1[0]].fst;
      ans *= b0;
      ans %= MOD;
      ans *= modInv(a1);
      ans %= MOD;
    } else {
      if (zcnt > 0) {
        ans = 0;
      } else {
        ans = 1;
        REP(i, K) {
          ans *= a[a.size() - 1 - 1 - i].fst;
          ans %= MOD;
        }
        DUMP(ans);
        ans *= -1;
        ans += MOD;
      }
    }
    print(ans);
  }
}