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;
#define ll long long
#define ld long double
#define oo 0x3f3f3f3f

int main()
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n,k;
    cin>>n>>k;
    vector<array<int,2>>A;
    int posCnt = 0;

    for(int i=0; i<n; i++)
    {
        int x;
        cin>>x;
        posCnt += x >=0;
        A.push_back({abs(x), x < 0 ? -1 : x==0 ? 0 : 1});
    }
    sort(A.rbegin(),A.rend());

    int cnt = 0;

    ll lastNeg = 1;
    ll lastPos = 1;

    ll res = 1;
    ll mod = 1e9+7;

    for(int i=0; i<k; i++)
        if(A[i][1] < 0)
        {
            res = (res*lastNeg)%mod;
            lastNeg=-A[i][0],cnt++;
        }
        else
        {
            res = (res*lastPos)%mod;
            lastPos = A[i][0];
        }

    if(cnt&1) //gotta get rid of minus
    {
        ll nxtNeg = 1;
        ll nxtPos = -1;

        for(int i=k; i<n; i++)
            if(A[i][1] < 0)
        {
            nxtNeg=-A[i][0];
            break;
        }

        for(int i=k; i<n; i++)
            if(A[i][1] > 0)
        {
            nxtPos=A[i][0];
            break;
        }

        //case 1: get rid of lastPositive and add nxtNeg instead
        //case 2: get rid of lastNegative and add nxtPos instead

        if(nxtNeg < 0 && nxtPos >=0)
        {
                if(lastNeg * nxtNeg >= lastPos * nxtPos)
            {
                res = (res * lastNeg)%mod;
                res = (res * nxtNeg)%mod;
            }
            else
            {
                res = (res * lastPos)%mod;
                res = (res * nxtPos)%mod;
            }
        }
        else if(nxtNeg < 0)
        {
            res = (res * lastNeg)%mod;
            res = (res * nxtNeg)%mod;
        }
        else if(nxtPos >= 0)
        {
            res = (res * lastPos)%mod;
            res = (res * nxtPos)%mod;
        }
    }
    else
    {
        res = (res * lastPos)%mod;
        res = (res * lastNeg)%mod;
    }

    if(res < 0)res+=mod;


    if(posCnt == 0) //all negative
    {
        res = 1;
        reverse(A.begin(),A.end());
        for(int i=0; i<k; i++)
            res = (res * (-A[i][0]))%mod;
    }
    if(res < 0)res+=mod;

    cout<<res;
}