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>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
#define N 1000000007
int main() {
  int64_t n, k, pos = 0, neg = 0, zero = 0, flag = 0, negct = 0,
                posct = 0, zeroct = 0;
  cin >> n >> k;
  boost::multiprecision::cpp_int ans=1;
  int64_t a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] > 0)
      pos++;
    else if (a[i] < 0)
      neg++;
    else
      zero++;
  }
  sort(a, a + n);
  if (k % 2 == 1 && pos + zero == 0)
    flag = 1;

  if (flag == 1) {
    while (k > 0) {
      ans *= a[neg - 1 - negct];
      ans = (ans + N) % N;
      negct++;
      k--;
    }
    cout << ans;
    return 0;
  }
  while (k >= 2) {
    if (a[0 + negct] * a[1 + negct] > a[n - 1 - posct] * a[n - 2 - posct]) {
      ans *= a[0 + negct]*a[1 + negct];
      ans = (ans + N) % N;
      negct += 2;
      k -= 2;
    } else {
      ans *= a[n - 1 - posct]*a[n - 2 - posct];
      ans = (ans + N) % N;
      posct += 2;
      k -= 2;
    }
  }
  if (k == 1) {
    ans *= a[n - 1 - posct];
    ans = (ans + N) % N;
    while(ans<0){
      ans+=N;
    }
  }
  cout << ans;
  return 0;
}