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>

using namespace std;

#define PB push_back
#define f first
#define s second
#define what_is(x) cerr << #x << " is " << x << endl;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;

const ll MOD = 1000000007;
const int N = 200005;
int n, k;
vector<ll> a[2];

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> k;
    for(int i=0; i<n; i++)
    {
        ll x;
        cin >> x;
        a[x<0].PB(x);
    }
    if(a[0].size())
        sort(a[0].begin(), a[0].end());
    if(a[1].size())
        sort(a[1].begin(), a[1].end());
    int siz0 = a[0].size();
    int siz1 = a[1].size();
    //what_is(siz0);
    //what_is(siz1);
    int ind0 = siz0-k, ind1 = 0;
    //take all positive
    if((siz1 == 0) || (siz1 > 0 && siz0 >= k && abs(a[1][0]) < a[0][siz0-k]))
    {
        ind0 = siz0-k;
        ind1 = 0;
    }
    else
    {
        for(int i=2; i<=min(k, (int)siz1); i+=2)
        {
            if(k - i > siz0)
                continue;
            int ind = siz0-k+i;
            //what_is(i);
            //what_is(ind);
            int smallest = i-1;
            if(ind == 0)
            {
                ind1 = i;
                ind0 = ind;
            }

            if(ind >= 2 && (a[0][ind-1] * a[0][ind-2]) <= (a[1][smallest] * a[1][smallest-1]))
            {
                ind1 = i;
                ind0 = ind;
            }
        }
        //all negative

    }
    ll ans = 1LL;
    //what_is(ind0);
    //what_is(ind1);
    if(ind0 < 0)
    {
        if(k%2 == 1)
        {
            for(int i=siz1-k; i<siz1; i++)
            {
                ans = (ans * (MOD + a[1][i]) % MOD) % MOD;
            }
        }
        else
        {
            for(int i=0; i<k; i++)
            {
                ans = (ans * (MOD + a[1][i]) % MOD) % MOD;
            }
        }
    }
    else
    {
        for(int i=ind0; i<siz0; i++)
        {
            ans = (ans * a[0][i]) % MOD;
        }
        for(int i=0; i<ind1; i++)
        {
            ans = (ans * (MOD + a[1][i]) % MOD) % MOD;
        }
    }
    cout << ans;
    return 0;
}