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: Runtime Error)

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define whole(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bit(n) (1LL<<(n))
#define F first
#define S second
// functions
template <typename T> void unique(T& c){c.erase(std::unique(c.begin(), c.end()), c.end());}
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 <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << "(" << p.first << ", " << p.second << ")"; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 60;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(nullptr); cout << fixed << setprecision(20);}
} fast_io_;

template <typename T>
T power(T a, T n, T mod) {
  T res = 1;
  T tmp = n;
  T curr = a;
  while(tmp){
    if(tmp % 2 == 1){
      res = (T)(res * curr % mod);
    }
    curr = (T)(curr * curr % mod);
    tmp >>= 1;
  }

  return res;
}

template<typename T>
T extgcd(T a, T b, T &x, T &y){ 
  T d = a;
  if(b != 0){
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  }else{
    x = 1, y = 0;
  }
  return d;
}

template <typename T>
T modinv(T a, T m){
  long long x = 0, y = 0;
  extgcd<long long>(a, m, x, y);
  x %= m;
  if(x < 0)x += m;
  return x;
}


int main(int argc, char const* argv[])
{
  int n, k; cin >> n >> k;
  vector<ll> pos, neg;
  vector<ll> a(n); cin >> a;
  int zero = 0;
  rep(i, n){
    if(a[i] > 0)pos.pb(a[i]);
    else if(a[i] < 0)neg.pb(a[i]);
    else zero++;
  }
  sort(whole(pos));
  sort(whole(neg));
  if(k == n){
    ll res = a[0];
    FOR(i, 1, n)res = res * a[i] % mod;
    if(res < 0)res += mod;
    cout << res << endl;
    return 0;
  }
  // minus
  if(sz(neg) == n && k % 2 == 1){
    reverse(whole(neg));
    ll res = neg[0];
    FOR(i, 1, k)res = res * neg[i] % mod;
    if(res < 0)res += mod;
    cout << res << endl;
    return 0;
  }
  // zero
  if(k > sz(pos) + sz(neg)){
    cout << 0 << endl;
    return 0;
  }
  ll res = 0;
  ll curr = 1;
  int j = 0, c = 0;
  reverse(whole(pos));
  if(k & 1){
    curr = pos[0];
    j++;
    c++;
  }
  for(; j + 1 < k && j + 1 < sz(pos); j += 2){
    curr = curr * pos[j] % mod;
    curr = curr * pos[j+1] % mod;
    c += 2;
  }
  int npos = c;
  j = 0;
  while(c < k && j + 1 < sz(neg)){
    curr = curr * neg[j] % mod;
    curr = curr * neg[j+1] % mod;
    c += 2;
    j += 2;
  }
  res = curr;
  while(j + 1 < sz(neg) && npos > 1){
    ll tmp1 = neg[j] * neg[j+1];
    ll tmp2 = pos[npos-1] * pos[npos-2];
    if(tmp1 >= tmp2){
      curr = curr * neg[j] % mod;
      curr = curr * neg[j+1] % mod;
      curr = curr * modinv<ll>(pos[npos-1], mod) % mod;
      curr = curr * modinv<ll>(pos[npos-2], mod) % mod;
      npos -= 2;
      j += 2;
      res = max(res, curr);
    }else{
      break;
    }
  }

  cout << res << endl;
  return 0;
}