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;
typedef long long ll;
#define all(x) (x).begin(), (x).end()
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vector<bool>> vvb;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
void debugF(ostream &os) {os << endl;}
template <typename Head, typename... Tail>
void debugF(ostream &os, Head H, Tail... T) { os << " " << H; debugF(os, T...); }
#define debug(...) cout << "[" << #__VA_ARGS__ << "]:", debugF(cout, __VA_ARGS__)
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); os << (i>0 ? " ":"") << v[i++]); return os;}

ll inf = 2e18; 
ll mod = 1e9 + 7;

void resposta(vector<pll> &a, int k){
    ll ans = 1;
    int i;
    for(i=0;i<k;i++){
        ans *= a[i].second;
        ans %= mod;
        if(ans < 0) ans += mod;
    }
    cout << ans << "\n";
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k, i, j;
    cin >> n >> k;
    vector<pll> a(n);
    bool hasP = false;
    for(i=0;i<n;i++){
        ll x; cin >> x;
        a[i] = {abs(x), x};
        if(x >= 0) hasP  = true;
    }
    sort(all(a));
    ll ans = 1;
    if(k%2 == 1 && (!hasP)){
        resposta(a, k);
        return 0;
    }
    reverse(all(a));
    // for(i=0;i<n;i++){
    //     cout << a[i].first << " " << a[i].second << "\n";
    // }
    ll minP = inf, minN = inf;
    int imp, imn;
    int cntN = 0;
    for(i=0;i<k;i++){
        if(a[i].second < 0){
            cntN++;
            minN = a[i].second;
            imn = i;
        }else{
            minP = a[i].second;
            imp = i;
        }
    }
    ll nxtP = inf, nxtN = inf;
    for(i=k; i < n;i++){
        if(a[i].second < 0){
            if(nxtN == inf) nxtN = a[i].second;
        }else{
            if(nxtP == inf) nxtP = a[i].second;
        }
    }
    if(cntN%2 == 0){
        resposta(a, k);
        return 0;
    }
    if(minP == inf){
        a[imn] = {abs(nxtP), nxtP};
        resposta(a, k);
        return 0;
    }
    assert(minP != inf && minN != inf);
    ll op1, op2;
    if(nxtN == inf){
        op1 = -inf;
    }else{
        op1 = minN * nxtN;
    }
    if(nxtP == inf){
        op2 = -inf;
    }else{
        op2 = nxtP * minP;
    }
    if(op1 > op2){
        a[imp] = {abs(nxtN), nxtN};
    }else{
        a[imn] = {nxtP, nxtP};
    }

    resposta(a, k);
    return 0;
}