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: Accepted)

#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <numeric>
#include <string>
#include <iostream>
#include <fstream>
#include <algorithm>
#include <assert.h>
#include <ctype.h>
#include <limits.h>
#include <math.h>
#include <cmath>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <bitset>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <sstream>

using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-11;
typedef long long ll;
const ll MOD = 1000000007;
const int N = 100000 + 7;
typedef unsigned long long ull;

//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
void fast() {
    std::ios_base::sync_with_stdio(0);
    cin.tie(0);
}

/*                What if i lose? only losers think like that. But the answer is you will lose rate.                */


ll n, k, x;
vector<ll> arr;

int main() {
#ifndef ONLINE_JUDGE
    freopen("Input.txt", "r", stdin);
    freopen("Output.txt", "w", stdout);
#endif
    fast();

    cin >> n >> k;
    arr.resize(n);
    for (int i = 0; i < n; i++) {
        cin >> arr[i];
    }
    sort(arr.begin(),arr.end());

    if(arr[arr.size()-1]==0&&k&1)
        return puts("0"),0;

    ll ans = 1;
    if(k&1 && arr[arr.size()-1]<0){
        for(int i=arr.size()-1;k;i--,k--){
            ans*=arr[i];
            ans=(ans%MOD+MOD)%MOD;
        }
        return cout<<ans<<endl,0;
    }

    int ptr1=0,ptr2=arr.size()-1;
    if(k&1){
        ans=arr[arr.size()-1];
        k--;
        ptr2--;
    }

    while(k){
        if(arr[ptr1]*arr[ptr1+1]>arr[ptr2]*arr[ptr2-1]){
            k-=2;
            ans*=arr[ptr1];
            ans=(ans%MOD+MOD)%MOD;
            ans*=arr[ptr1+1];
            ans=(ans%MOD+MOD)%MOD;
            ptr1+=2;
        }else{
            k-=2;
            ans*=arr[ptr2];
            ans=(ans%MOD+MOD)%MOD;
            ans*=arr[ptr2-1];
            ans=(ans%MOD+MOD)%MOD;
            ptr2-=2;
        }
    }
    cout<<ans<<endl;

    return 0;
}