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;
int main(){
  int N, K; cin>>N>>K;
  long long M = 1e9 + 7;
  vector<long long> pos, neg, l(1), r(1);
  long long zero = 0;
  for(int i = 0; i < N; ++i){
    long long a; cin>>a;
    
    if(a > 0) pos.push_back(a);
    else if(a == 0)
      zero = 1;
    else
    {
      neg.push_back(a);
    }
    
  } 
  l[0] = 1;r[0] = 1;
  long long ans = -2e18;
  
  sort(pos.begin(), pos.end(), greater<long long>());
  sort(neg.begin(), neg.end());
  for(auto &n : neg) n += M;
  for(int i = 0; i < N; ++i) {
    if(i < pos.size())
      l.push_back(pos[i] * l[i] % M);
    if(i < neg.size())
      r.push_back(neg[i] * r[i] % M);
  }
  if(pos.size() == 0 && K % 2) {
    if(zero) cout<<0<<endl;
    else {
      long long p = 1;
      for(int i = 0; i < K; ++i) {
        p = p * neg[neg.size() - 1 - i] % M;
      }
      cout<<p<<endl;
    }
    return 0;
  }
  if(zero) l.push_back(0);
  for(int i = 0; i <= K; i++) {
    if((K - i) % 2 == 0) {
      if(i < l.size() && K - i < r.size())
        ans = max(ans, l[i] * r[K - i] % M);
    }
  }
  cout<<(ans + M) % M<<endl;
  return 0;
}