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;
const int N=2e5+5,mod=1e9+7;
typedef long long ll;
int n,m,k,a[N],b[N],d[N],z;
long double aa[N],bb[N];
ll ma[N],mb[N];
int main()
{
    int s;scanf("%d%d",&s,&k);
    while(s--)
    {
        int x;scanf("%d",&x);
        if(x>0) a[++n]=x;
        else if(x<0) b[++m]=-x;
        else z++;
    }
    if(n) sort(a+1,a+1+n,greater<int>());
    if(m) sort(b+1,b+1+m,greater<int>());
    ma[0]=mb[0]=1;
    for(int i=1;i<=n;i++) aa[i]=aa[i-1]+log2((long double)a[i]),ma[i]=ma[i-1]*a[i]%mod;
    for(int i=1;i<=m;i++) bb[i]=bb[i-1]+log2((long double)b[i]),mb[i]=mb[i-1]*b[i]%mod;
    if(n+m<k){printf("0\n");return 0;}
    if(n+m==k)
    {
        if(m&1)
        {
            if(z) printf("0\n");
            else
                printf("%lld\n",(ma[n]*mb[m]%mod*-1+mod)%mod);
        }
        else
        {
            printf("%lld\n",(ma[n]*mb[m]%mod+mod)%mod);
        }
        return 0;
    }
    if(!n)
    {
        if(k&1)
        {
            if(z) printf("0\n");
            else
            {
                reverse(b+1,b+1+m);
                ll ans=1;
                for(int i=1;i<=k;i++) ans=ans*b[i]%mod;
                printf("%lld\n",(ans*-1+mod)%mod);
            }
        }
        else
        {
            printf("%lld\n",(mb[k]+mod)%mod);
        }
        return 0;
    }
    assert(0);
}