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>
#define REP(i, n) for (ll i = 0; i < n; ++i)
#define FOR(i, b, n) for (ll i = b; i < n; ++i)

using namespace std;
using ll = long long;
using P = pair<int, int>;
const ll MOD = 1000000007;

void solve(ll N, ll K, vector<ll> A) {
  sort(A.begin(), A.end());

  if (A.back() < 0 && K % 2) {
    ll ans(1);
    REP(i, K) {
      ans *= A[N - 1 - i];
      ans %= MOD;
    }
    if (ans < 0)
      ans += MOD;
    cout << ans << endl;
  } else {
    ll ans(1);
    ll L(0), R(N-1);
    while (K > 0) {
      if (K >= 2 && (A[L] * A[L+1]  > A[R] * A[R - 1])) {
	ans *= A[L];
	ans %= MOD;
	ans *= A[L+1];
	ans %= MOD;
	L += 2;
	K -= 2;
      } else {
	ans *= A[R];
	ans %= MOD;
	R--;
	K--;
      }
    }
    cout << ans << endl;
  }
}

int main(){
  // int 3E4  long 2E9  ll 9E18
  cin.tie(0);
  ios::sync_with_stdio(false);

  ll N;
  cin >> N;
  ll K;
  cin >> K;
  vector<ll> A(N);
  REP(i, N) {
    cin >> A[i];
  }

  solve(N, K, move(A));

  return 0;
}