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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<time.h>
#include<random>
#include<array>
using namespace std;
using ll = long long;
using ull = unsigned long long;
#define rep(i, a, b) for(int i = a; i < b; i++)
#define rrep(i, a, b) for(int i = b - 1; i >= a; i--)
#define ALL(a) a.begin(), a.end()
using pii = pair<int,int>;
using piii = pair<pii,int>;
using pll = pair<long long, long long>;
using plll = pair<pll, long long>;
// #pragma GCC optimize("Ofast")
#define pcnt __builtin_popcount
#define buli(x) __builtin_popcountll(x)
#define pb push_back
#define mp make_pair
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define isSquare(x) (sqrt(x)*sqrt(x) == x)
template<class T>inline bool chmax(T &a, const T &b) {if(a<b){a = b; return 1;} return 0; };
template<class T>inline bool chmin(T &a, const T &b) {if(a>b){a = b; return 1;} return 0; };
inline void in(void){return;}
template <typename First, typename... Rest> void in(First& first, Rest&... rest){cin >> first;in(rest...);return;}
inline void out(void){cout << "\n";return;}
template <typename First, typename... Rest> void out(First first, Rest... rest){cout << first << " ";out(rest...);return;}
const double EPS = 1e-9;
const int mod = 1e9 + 7;
// const int mod = 998244353;
const int INF = 1e9;
const long long INFLL = 1e18;
void iosetup() {
    cin.tie(nullptr);ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    cerr << fixed << setprecision(10);
}
template< typename T1, typename T2 >
ostream &operator<<(ostream &os, const pair< T1, T2 >& p) {
    os << p.first << " " << p.second;
    return os;
}
template< typename T1, typename T2 >
istream &operator>>(istream &is, pair< T1, T2 > &p) {
    is >> p.first >> p.second;
    return is;
}
template< typename T >
ostream &operator<<(ostream &os, const vector< T > &v) {
    for(int i = 0; i < (int) v.size(); i++) {
        os << v[i] << (i + 1 != v.size() ? " " : "");
    }
    return os;
}
template< typename T >
istream &operator>>(istream &is, vector< T > &v) {
    for(T &in : v) is >> in;
    return is;
}
template<class T> vector<T> make_vec(size_t a) {return vector<T>(a); }
template<class T, class... Ts> auto make_vec(size_t a, Ts... ts){
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template<class S, class T> pair<S,T> operator+(const pair<S,T> &s, const pair<S, T>& t){return pair<S,T>(s.first+t.first, s.second+t.second);}
template<class S, class T> pair<S,T> operator-(const pair<S,T> &s, const pair<S, T>& t){return pair<S,T>(s.first-t.first, s.second-t.second);}
template<class S, class T> pair<S,T> operator*(const pair<S,T> &s, const S& t){return pair<S,T>(s.first*t, s.second*t);}
template <typename T> void Exit(T first){cout << first << endl;exit(0); };
template< int mod > struct ModInt {
    unsigned x; ModInt() : x(0) {}
    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    ModInt &operator+=(const ModInt &p) {if((x += p.x) >= mod) x -= mod;return *this;}
    ModInt &operator-=(const ModInt &p) {if((x += mod - p.x) >= mod) x -= mod;return *this;}
    ModInt &operator*=(const ModInt &p) {x = (int) (1LL * x * p.x % mod);return *this;}
    ModInt &operator/=(const ModInt &p) {*this *= p.inverse();return *this;}
    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
    bool operator==(const ModInt &p) const { return x == p.x; }
    bool operator!=(const ModInt &p) const { return x != p.x; }
    ModInt inverse() const {int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) { t = a / b; swap(a -= t * b, b); swap(u -= t * v, v); }return ModInt(u);}
    ModInt pow(int64_t n) const {ModInt ret(1), mul(x); while(n > 0) {if(n & 1) ret *= mul;mul *= mul;n >>= 1;}return ret;}
    friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x;}
    friend istream &operator>>(istream &is, ModInt &a) { int64_t t; is >> t; a = ModInt< mod >(t); return (is); }
    static int get_mod() { return mod; }
}; using modint = ModInt< mod >;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const pii dxy[4] = {pii(1,0), pii(0, 1), pii(-1, 0), pii(0, -1)};
bool range(int a, int b, int x){if(a <= x and x < b)return true;else return false;}
bool range(int a, int b, int c, int d, pii p){if(a <= p.first and p.first < b and c <= p.second and p.second < d) return true;else return false;}


int main(){
    iosetup();
    int n, k; cin >> n >> k;
    vector<ll> a(n); cin >> a;
    vector<ll> b, c;
    rep(i, 0, n){
        if(a[i] >= 0) b.emplace_back(a[i]);
        else if(a[i] < 0)c.emplace_back(a[i]);
        // else d.emplace_back(a[i]);
    }
    sort(ALL(b), greater<ll>());
    sort(ALL(c));
    int idx1 = 0, idx2 = 0;
    int cnt = 0;
    modint ans = 1;
    if(k == n){
        rep(i, 0, n) ans *= a[i];
        Exit(ans);
    }
    if(b.empty()){
        sort(ALL(c), greater<ll>());
        rep(i, 0, k) ans *= c[i];
        Exit(ans);
    }
    if(c.size()%2==1) c.pop_back();
    while(cnt < k){
        if(k - cnt == 1){
            if(idx1 >= b.size()) ans *= c[idx2++];
            else ans *= b[idx1++];
            cnt++;
        }else if(idx2 + 1 >= c.size()){
            ans *= b[idx1++]; cnt++;
        }else if(idx1 >= b.size()){
            ans *= c[idx2] * c[idx2 + 1] % mod; 
            cnt += 2; idx2 += 2;
        }else if(idx1 + 1 >= b.size()){
            ans += c[idx2] * c[idx2+1] % mod;
            idx2 += 2; cnt += 2;
        }else{
            if(b[idx1] * b[idx1+1] <= c[idx2] * c[idx2 + 1]){
                ans *= c[idx2] * c[idx2 + 1] % mod;
                idx2 += 2; cnt += 2;
            }else{
                ans *= b[idx1];
                idx1 += 1; cnt += 1;
            }
        }
    }
    cout << ans << endl;
    return 0;
}