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 rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

const int mod = 1000000007;
struct mint{
	ll x;
	mint(ll x = 0):x((x%mod+mod)%mod){}
	mint operator-() const{return mint(-x);}
	mint& operator+=(const mint a){
		if((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a){
		if((x += mod-a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a){
		(x *= a.x) %= mod;
		return *this;
	}
	mint operator+(const mint a) const{
		mint res(*this);
		return res+=a;
	}
	mint operator-(const mint a) const{
		mint res(*this);
		return res-=a;
	}
	mint operator*(const mint a) const{
		mint res(*this);
		return res*=a;
	}
	mint pow(ll t) const{
		if(!t) return 1;
		mint a = pow(t>>1);
		a *= a;
		if(t&1) a *= *this;
		return a;
	}
	mint inv() const{
		return pow(mod-2);
	}
	mint& operator/=(const mint a){
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const{
		mint res(*this);
		return res/=a;
	}
	bool operator==(const mint a){
		return x == a.x;
	}
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    vector<ll> a(n);
    vector<ll> plus, minus;
    rep(i,n){
        cin >> a[i];
        if(a[i] >= 0) plus.push_back(a[i]);
        else minus.push_back(a[i]);
    }
    int ps = plus.size(), ms = minus.size();
    bool canPositive = false;
    if(ps > 0){
        if(n == k){
            if(ms%2 == 0) canPositive = true;
            else canPositive = false;
        }
        else{
            canPositive = true;
        }
    }
    else{
        if(k%2 == 0) canPositive = true;
        else canPositive = false;
    }
    mint ans = 1;
    if(canPositive){
        sort(plus.begin(), plus.end());
        sort(minus.rbegin(), minus.rend());
        if(k%2 == 1){
            ans *= plus[ps-1];
            plus.pop_back();
            --ps;
        }
        vector<ll> positivePair;
        for(int i = 0; i < ps; i+=2){
            positivePair.push_back(plus[i] * plus[i+1]);
        }
        for(int j = 0; j < ms; j+=2){
            positivePair.push_back(minus[j] * minus[j+1]);
        }
        sort(positivePair.rbegin(), positivePair.rend());
        rep(i,k/2) ans *= positivePair[i];
    }
    else{
        vector<P> absA;
        rep(i,n){
            if(a[i] >= 0) absA.push_back(P(a[i], a[i]));
            else absA.push_back(P(-a[i], a[i]));
        }
        sort(absA.begin(), absA.end());
        rep(i,k) ans *= absA[i].second;
    }
    cout << ans.x << endl;
    return 0;
}