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>
using namespace std;

#define REP(i,m,n) for(int i=(m); i<(int)(n); i++)
#define RREP(i,m,n) for(int i=(int)(n-1); i>=m; i--)
#define rep(i,n) REP(i,0,n)
#define rrep(i,n) RREP(i,0,n)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define fi first
#define se second
#define debug(...) {cerr<<"[L"<<__LINE__<<"] "; _debug(__VA_ARGS__);}

template<typename T>
string join(const vector<T>&v, string del=", "){ stringstream s;
    for(auto x : v) s << del << x; return s.str().substr(del.size());
}
template<typename T>
ostream& operator<<(ostream& o, const vector<T>&v){
    if(v.size()) o << "[" << join(v) << "]"; return o;
}
template<typename T>
ostream& operator<<(ostream& o, const vector<vector<T> >&vv){
    int l = vv.size();
    if(l){ o<<endl; rep(i,l) o << (i==0 ? "[ " : ",\n  " ) << vv[i] << (i==l-1 ? " ]" : ""); }
    return o;
}
template<typename T1, typename T2>
ostream& operator<<(ostream& o, const pair<T1, T2>& p){
    return o << "(" << p.first << ", " << p.second << ")";
}
inline void _debug(){cerr<<endl;}
template<class First, class... Rest>
void _debug(const First& first, const Rest&... rest){cerr<<first<<" ";_debug(rest...);}

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;

const double PI = (1*acos(0.0));
const double EPS = 1e-9;
const int INF = 0x3f3f3f3f;
const ll INFL = 0x3f3f3f3f3f3f3f3fLL;
const ll mod = 1e9 + 7;

inline void finput(string filename) {
    freopen(filename.c_str(), "r", stdin);
}

int main(){
    ios_base::sync_with_stdio(0);
    // finput("./input");
    int n,k; cin >> n >> k;
    vl P, M;
    rep(i,n){
        ll a; cin >> a;
        if(a > 0) P.push_back(a);
        else M.push_back(abs(a));
    }
    if(k == n){
        ll ans = 1;
        rep(i,P.size()){
            ans *= P[i];
            ans %= mod;
        }
        rep(i,M.size()){
            ans *= mod - M[i];
            ans %= mod;
        }
        cout << ans << endl;
        return 0;
    }
    if(k & 1 && P.size() == 0){
        ll ans = 1;
        rep(i,k){
            ans *= mod - M[i];
            ans %= mod;
        }
        cout << ans << endl;
        return 0;
    }
    sort(rall(P));
    sort(rall(M));

    rep(i,M.size()) M[i] = mod - M[i];
    vector<pll> PS(P.size()+1, pll(1, 0)), MS(M.size()+1, pll(1, 0));
    rep(i,P.size()){
        ll prod = PS[i].fi * P[i];
        PS[i+1].fi = prod % mod;
        PS[i+1].se = PS[i].se + (prod / mod);
    }
    rep(i,M.size()){
        ll prod = MS[i].fi * M[i];
        MS[i+1].fi = prod % mod;
        MS[i+1].se = MS[i].se + (prod / mod);
    }

    pll mx = pll(0, 0);
    ll ans = 0;
    for(ll i=0; i<=M.size(); i+=2){
        int j = k - i;
        if(j > P.size()) continue;
        pll p;
        ll prod = MS[i].fi * PS[j].fi;
        p.fi = MS[i].se + PS[j].se + (prod / mod);
        p.se = prod % mod;
        if(p.fi > mx.fi or (p.fi == mx.fi && p.se > mx.se)){
            ans = p.se;
        }
    }
    cout << ans << endl;
    return 0;
}