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() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<long long> all(n);
  for (int i = 0; i < n; i++) {
    cin >> all[i];
  }
  sort(all.begin(), all.end(), [&](int a, int b) {
    return abs(a) > abs(b);
  });
  int flag = 0, pi = -1, pj = -1, ni = -1, nj = -1, cnt = 0;
  for (int i = 0; i < n; i++) {
    cnt += (all[i] < 0);
    flag ^= (all[i] < 0 && i < k);
    if (all[i] < 0) {
      if (i >= k && nj == -1) nj = i;
      else if (i < k) ni = i;
    } else {
      if (i >= k && pj == -1) pj = i;
      else if (i < k) pi = i;
    }
  }
  if (flag && cnt == n) {
    reverse(all.begin(), all.end());
  } else if (flag) {
    if (pi  != -1 && nj != -1 && ni != -1 && pj != -1) {
      if (all[nj] / all[pi] > all[pj] / all[ni]) {
        swap(all[pi], all[nj]);
      } else {
        swap(all[ni], all[pj]);
      }
    } else if (pi  != -1 && nj != -1) {
      swap(all[pi], all[nj]);
    } else if (ni != -1 && pj != -1) {
      swap(all[ni], all[pj]);
    }
  }
  const long long mod = 1e9 + 7;
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    ans = ans * all[i] % mod;
    ans = (ans + mod) % mod;
  }
  cout << ans << endl;
  return 0;
}