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: Runtime Error)

#include <bits/stdc++.h>

#define FLASH ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define ll long long int
#define MAX 1e9
#define MIN -1e9
#define mod 1000000007
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define debug(x) cerr<<#x<<'='<<(x)<<endl;
#define mxn 300005
using namespace std;
ll domod(ll a, ll b){
    a = (a * b) % mod;
    a += mod;
    a %= mod;
    return a;
}
int main()
{
    FLASH
    ll n,k,ans = 1;
    cin >> n >> k;
    vector<ll> po,ne;
    for(int i = 0; i < n; i++){
        ll x;
        cin >> x;
        if(x > 0){
            po.pb(x);
        }
        else{
            ne.pb(x);
        }
    }
    ll nn = po.size(),mm = ne.size(),pos,neg;
    neg = 0,pos = nn;
    sort(all(po));
    reverse(all(po));
    sort(all(ne));
    if(k == n){
        for(int i = 0; i < po.size(); i++)ans = domod(ans,po[i]);
        for(int i = 0; i < ne.size(); i++)ans = domod(ans,ne[i]);
        cout << ans << '\n';
        return 0;
    }
    if(nn == 0){
        reverse(all(ne));
        for(int i = 0; i < k; i++)ans = domod(ans,ne[i]);
        cout << ans << '\n' ;
        return 0;
    }
    pos == min(nn,k);
    neg = k - nn;
    if(neg % 2 == 1){
        pos--;
        neg++;
    }
    while(neg + 1 <= mm - 1 && pos - 2 >= 0 && ne[neg] * ne[neg + 1]  > po[pos - 1] * po[pos - 2]){
        neg += 2;
        pos -= 2;
    }
    for(int i = 0; i < pos; i++)
        ans = domod(ans,po[i]);
    for(int i = 0; i < neg; i++){
        ans = domod(ans,ne[i]);
    }
    cout << ans << '\n' ;
    return 0;
}