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 <bits/stdc++.h>

using namespace std;
#define all(a)a.begin(),a.end()
using ll=long long;
const int INF = 1<<30;
const ll INFll =1LL<<62;
const int mod= int(1e9)+7;
using P = pair<ll,ll>;
using ld=long double;

// modint: mod 計算を int を扱うように扱える構造体
template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};
using mint = Fp<1000000007>;

int main(){
    ll n,k;cin>>n>>k;
    vector<ll>a(n);
    for (int i = 0; i < n; ++i)cin >> a[i];
    sort(all(a),[&](ll x,ll y){return abs(x)>abs(y);});
    mint ans=1;
        ll cnt = 0;
        ll lastm = -1;
        ll lastp = -1;
        ll fp = -1;
        ll fs = -1;
        if(k!=n) {
            for (int i = k; i < n; ++i) {
                if(a[i]<0){
                    fs=i;
                    break;
                }
            }
            for (int i = k; i <n; ++i) {
                if(a[i]>=0){
                    fp=i;
                    break;
                }
            }
        }
        for (int i = 0; i < k; ++i) {
            if (a[i] < 0) {
                cnt++;
                lastm = i;
            }
            else lastp=i;
        }
        if (cnt % 2 == 0) {
            for (int i = 0; i < k; ++i) {
                ans *= a[i];
            }//ok
        } else {
            if(lastm>=0&&fp>=0&&lastp>=0&&fs>=0){
                if(abs(a[lastm])*abs(a[fs])>abs(a[lastp])*abs(a[fp])){
                    //pぬきm
                    for (int i = 0; i < k; ++i) {
                        if(i!=lastp)ans*=a[i];
                    }
                    ans*=a[fs];
                }
                else {
                    for (int i = 0; i < k; ++i) {
                        if(i!=lastm)ans*=a[i];
                    }
                    ans*=a[fp];
                }
            }
            else if(lastm>=0&&fp>=0){
                for (int i = 0; i < k; ++i) {
                    if(i!=lastm)ans*=a[i];
                }
                ans*=a[fp];
            }
            else if(lastp>=0&&fs>=0){
                for (int i = 0; i < k; ++i) {
                    if(i!=lastp)ans*=a[i];
                }
                ans*=a[fs];
            }
            else {
                reverse(all(a));
                for (int i = 0; i < k; ++i) {
                    ans*=a[i];
                }
            }
        }
    cout <<ans <<endl;
}