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: Accepted)

#include <bits/stdc++.h>
//#include <boost/multiprecision/cpp_int.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("O3")
#define REP(i,n) for(int i=0;i<n;i++)
#define REPP(i,n) for(int i=1;i<=n;i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define EPS (1e-9)
#define INF (1e17)
#define PI (acos(-1))
//const double PI = acos(-1);
//const double EPS = 1e-15;
//long long INF=(long long)1E17;
#define i_7 (long long)(1e9+7)
//#define i_7 998'244'353
long mod(long a){
    long long c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
long long po(long a, long b){
    if(b==0){
        return 1;
    }
    
    long long z = po(a,b/2);
    z = mod(z*z);
    if(b%2!=0){
        z = mod(a*z);
    }
    return z;
}
 
bool prime_(int n){
  if(n==1){
    return false;
  }else if(n==2){
    return true;
  }else{
    for(int i=2;i<=std::sqrt(n);i++){
      if(n%i==0){
        return false;
      }
    }
    return true;
  }
}
 
long long gcd_(long long a, long long b){
  if(a<b){
    std::swap(a,b);
  }
  if(a%b==0){
    return b;
  }else{
    return gcd_(b,a%b);
  }
}
 
long long lcm_(long long x, long long y){
  return (x/gcd_(x,y))*y;
}
 
//using namespace std;
//using namespace boost::multiprecision;
//using namespace __gnu_pbds;

int main(){
  using namespace std;
  int n, k;
  cin>>n>>k;
  vector<long long> a(n);
  bool all_minus = true;
  REP(i, n){
    cin>>a[i];
    if(a[i] >= 0) all_minus = false;
  }
  auto cmp = [](long long x, long long y) -> bool{
    return abs(x) > abs(y);
  };
  sort(ALL(a), cmp);
  long long ans = 1;
  if(k == n){
    REP(i, k){
      ans *= a[i];
      ans = mod(ans);
    }
    cout << ans << endl;
    return 0;
  }
  if(all_minus && k % 2 == 1){
    reverse(ALL(a));
    REP(i, k){
      ans *= a[i];
      ans = mod(ans);
    }
    cout << ans << endl;
    return 0;
  }
  
  queue<long long> s;
  deque<long long> pos;
  deque<long long> neg;
  REP(i, n){
    if(a[i] >= 0LL){
      pos.push_back(a[i]);
    }else{
      neg.push_back(a[i]);
    }
  }
  while((int)s.size() < k){
    if((int)s.size() == (k - 1)){
      long long p1 = pos.front();
      pos.pop_front();
      s.push(p1);
      break;
    }
    if((int)neg.size() < 2){
      long long p1 = pos.front();
      pos.pop_front();
      s.push(p1);
    }else if(pos.empty()){
      long long n1 = neg.front();
      neg.pop_front();
      long long n2 = neg.front();
      neg.pop_front();
      s.push(n1);
      s.push(n2);
    }else if((int)pos.size() < 2){
      long long p1 = pos.front();
      pos.pop_front();
      long long n1 = neg.front();
      neg.pop_front();
      long long n2 = neg.front();
      neg.pop_front();
      if(p1 >= n1 * n2){
        s.push(p1);
        neg.push_front(n2);
        neg.push_front(n1);
      }else{
        s.push(n1);
        s.push(n2);
        pos.push_front(p1);
      }
    }else{
      long long p1 = pos.front();
      pos.pop_front();
      long long p2 = pos.front();
      pos.pop_front();
      long long n1 = neg.front();
      neg.pop_front();
      long long n2 = neg.front();
      neg.pop_front();
      if(p1 * p2 >= n1 * n2){
        s.push(p1);
        pos.push_front(p2);
        neg.push_front(n2);
        neg.push_front(n1);
      }else{
        s.push(n1);
        s.push(n2);
        pos.push_front(p2);
        pos.push_front(p1);
      }
    }
  }

  while(!s.empty()){
    long long x = s.front();
    s.pop();
    ans *= x;
    ans = mod(ans);
  }
  cout << ans << endl;
  return 0;
}