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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
using ll = long long;

const ll mod = 1e9 + 7;

int main() {
  int N, K;
  cin >> N >> K;
  vector<int> A(N), p, n;
  rep(i, N) {
    cin >> A[i];
    if (A[i] >= 0) p.push_back(A[i]);
    else n.push_back(A[i]);
  }

  int pn = p.size();
  int nn = n.size();
  bool ok = false;
  if (pn > 0) {
    if (N == K) ok = (nn % 2 == 0);
    else ok = true;
  }
  else {
    ok = (K % 2 == 0);
  }

  ll ans = 1;
  if (ok) {
    sort(p.begin(), p.end());
    sort(n.rbegin(), n.rend());
    if (K % 2 == 1) {
      ans *= p.back();
      p.pop_back();
    }
    vector<ll> t;
    while (p.size() >= 2) {
      ll x = p.back();
      p.pop_back();
      x *= p.back();
      p.pop_back();
      t.push_back(x);
    }
    while (n.size() >= 2) {
      ll x = n.back();
      n.pop_back();
      x *= n.back();
      n.pop_back();
      t.push_back(x);
    }
    sort(t.rbegin(), t.rend());
    rep(i, K / 2) {
      ans = ((ans % mod) * (t[i] % mod)) % mod;
    }
  }
  else {
    sort(A.begin(), A.end(), [](int x, int y) {
      return abs(x) < abs(y); 
    });
    rep(i, K) {
      ans = (ans * A[i]) % mod;
      if (ans < 0) ans += mod;
    }
  }

  cout << ans << endl;
  return 0;
}