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>
#define dalao main
#define debug(x) {cerr<<#x<<" = "<<(x)<<endl;}
#define ll long long
#define MOD 1000000007
using namespace std;

#define int ll
ll ans=1;
ll zheng[200005],fu[200005],jdz[200005];
int n,k,temp;
int zs=0,fs=0,zero=0;//正数和负数和0个数
#define int int
int dalao()
{
    #define int ll
    cin>>n>>k;
    for(int i=1;i<=n;i++)
    {
        cin>>temp;
        if(temp>0)
            zheng[++zs]=temp;
        else if(temp<0)
            fu[++fs]=temp;
        else
            zero++;
    }

    sort(zheng+1,zheng+zs+1);
    reverse(zheng+1,zheng+zs+1);
    sort(fu+1,fu+fs+1);

    if(k>zs+fs)
    {
        cout<<0;
        return 0;
    }
    if(zs==0)//都是负数
    {
        if(k%2==0)//偶数个负数相乘，取绝对值最大的k个
        {
            for(int i=1;i<=k;i++)
                ans=(ans%MOD) * (jdz[i]%MOD) % MOD;
            cout<<ans%MOD;
            return 0;
        }
        else//奇数个负数相乘，如果有0直接输出0，否则取绝对值最小的n个
        {
            if(zero>0)
            {
                cout<<0;
                return 0;
            }
            for(int i=fs;i>=fs-k+1;i--)
                ans=(ans%MOD) * ((fu[i])%MOD) % MOD;
            cout<<(MOD+ans)%MOD;
            return 0;
        }
    }

    //有正有负或都是正数
    int nowz=1,nowf=1,f=1;
    bool fushusile=false;
    for(int i=1;i<=k;i++)
    {
        //debug(f)debug(nowz)debug(nowf);
        if(nowf==fs && f==1)
            fushusile=true;
        if(nowf==fs && f==-1)
        {
            ans=(ans%MOD) * (fu[nowf]%MOD) % MOD;
            fushusile=true;
            continue;
        }


        if(fushusile || (zheng[nowz] > -fu[nowf]))
        {
            ans=(ans%MOD) * (zheng[nowz]%MOD) % MOD;
            nowz++;
        }
        else
        {
            ans=(ans%MOD) * (fu[nowf]%MOD) % MOD;
            nowf++;
            f*=-1;
        }
        //debug(ans)
    }
    cout<<ans%MOD;

    return 0;
}