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 pb push_back
#define mp make_pair
#define deb(a,b,c) cout<<#a<<" = "<<(a)<<", "<<#b<<" = "<<(b)<<", "<<#c<<" = "<<(c)<<endl;

typedef long long int ll;
typedef string str;

ll power(ll x, ll y, ll p) 
{ 
    ll res = 1;     
  
    x = x % p;   
    while (y) 
    { 
        if (y & 1) 
            res = (res*x) % p; 
        y = y>>1;  
        x = (x*x) % p; 
    } 
    return res; 
} 
 
ll nCrModP(ll n, ll r, ll p) 
{ 
    if (!r) 
      return 1; 
  
	ll fac[n+1]; 
   	fac[0] = 1; 
    for (ll i=1 ; i<=n; i++) 
        fac[i] = fac[i-1]*i%p; 
  
    return (fac[n]* power(fac[r], p-2,p) % p * power(fac[n-r], p-2,p) % p) % p; 
}

ll gcd(ll a, ll b){ 
    if (a == 0) 
        return b;  
    return gcd(b % a, a);
}

void fast(){
	ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}

int main ()
{
	
	ll t,n,m,k=0,i,j,x,y,p=pow(10,9)+7;
	cin >> n >> k;
    vector <ll> po,ne;
    for(i=0;i<n;i++)
    {
        cin >> m;
        if(m>=0)
            po.pb(m);
        else
            ne.pb(m);
    }
    sort(po.begin(),po.end());
    sort(ne.begin(),ne.end());
    x=po.size();
    y=ne.size();
    if(y==n)
    {
        t=1;
        for(i=n-1,j=0;j<k;i--,j++)
        {
            t*=ne[i];
            t=(t+p)%p;
            t%=p;
        }
        cout << (t+p)%p << "\n";
        return 0;
    }
    if(k%2)
    {
        t=po[x-1]%p;
        i=x-2;
        j=0;
        ll cnt=1;
        while(i>0 && j<y-1 && cnt <k)
        {
            ll a=po[i]*po[i-1],b=ne[j]*ne[j+1];
            if(a>b)
            {
                i-=2;
                a%=p;
                t=(a*t)%p;
                cnt+=2;
            }
            else
            {
                j+=2;
                b%=p;
                t=(b*t)%p;
                cnt+=2;
            }
        }
        while(i>0 && cnt<k)
        {
            ll a=po[i]*po[i-1];
            {
                i-=2;
                a%=p;
                t=(a*t)%p;
                cnt+=2;
            }

        }
        while(j<y-1 && cnt <k)
        {
            ll b=ne[j]*ne[j+1];
            {
                j+=2;
                b%=p;
                t=(b*t)%p;
                cnt+=2;
            }
        }
    }
    else
    {
        t=1;
        ll cnt=0;
        i=x-1;
        j=0;
        while(i>0 && j<y-1 && cnt <k)
        {
            ll a=po[i]*po[i-1],b=ne[j]*ne[j+1];
            if(a>b)
            {
                i-=2;
                a%=p;
                t=(a*t)%p;
                cnt+=2;
            }
            else
            {
                j+=2;
                b%=p;
                t=(b*t)%p;
                cnt+=2;
            }
        }
        while(i>0 && cnt<k)
        {
            ll a=po[i]*po[i-1];
            {
                i-=2;
                a%=p;
                t=(a*t)%p;
                cnt+=2;
            }

        }
        while(j<y-1 && cnt <k)
        {
            ll b=ne[j]*ne[j+1];
            {
                j+=2;
                b%=p;
                t=(b*t)%p;
                cnt+=2;
            }
        }
    }
    cout << t%p << "\n";
	return 0;
}