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)

// wrong answer
#include <bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 2e5 + 1;
const int inf = 1e9 + 7;
ll n,k,a[N],sport[N];
vector <ll> small,big;
ll power(ll x,ll y,ll mod){
    if (y == 1) return x % mod;
    if (y == 0) return 1;
    ll temp = power(x,y/2,mod) % mod;
    temp = (temp * temp) % mod;
    if (y & 1) return (temp * (x % mod)) % mod; return temp;
}
int main(){
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    /*sort(a + 1,a + 1 + n);
    a[0] = 1;
    //for (int i = 1; i <= n; i++) cout << a[i] << " " ; cout << "\n";
    for (int i = 1; i <= n; i++)
        if (a[i] < 0) sport[i] = sport[i - 1] + 1; else sport[i] = sport[i - 1];
    for (int i = 1; i <= n; i++) a[i] = (a[i - 1] * abs(a[i])) % inf;
    //for (int i = 1; i <= n; i++) cout << a[i] << " " ; cout << "\n";
    //for (int i = 1; i <= n; i++) cout << sport[i] << " "; cout << "\n";
    ll ans = -1e18;
    for (int x = 0; x <= k; x++){
        ll mul = (a[x] * (power(a[n - (k - x)],inf - 2,inf) * a[n]) % inf) % inf;
        ll gr = sport[x] + sport[n] - sport[n - (k - x)];
        if (gr % 2 == 1) mul = - mul;
        if (ans < mul)
            ans = max(ans,(mul + inf) % inf);
        //cout << x << " " << n - (k - x) << " " << mul << " " << gr << "\n";
    }
    cout << ans;*/
    for (int i = 1; i <= n; i++){
        if (a[i] < 0) small.push_back(a[i]);
        if (a[i] >= 0) big.push_back(a[i]);
    }
    sort(small.begin(),small.end());
    sort(big.begin(),big.end(),greater<ll>());
    int i = 0 , j = 0;
    ll ans = 1;
    while (k > 0){
        if ((int)small.size() - i - 1 > 0 && (int)big.size() - j > 0){
            if (big[j] < small[i] * small[i + 1] && k > 1){
                ans = (ans * ((small[i] + inf) % inf) * (small[i + 1] + inf) % inf) % inf;
                //cout << " -- " << small[i] << " " << small[i + 1] << "\n";
                i = i + 2;
                k -= 2;
            }
            else{
                ans = (ans * big[j]) % inf;
                //cout << " ++ " << big[j] << "\n";
                j++; k--;
            }
        }
        else
        if ((int)big.size() - j > 0){
            ans = (ans * big[j]) % inf;
            //cout << " + " << big[j] << "\n";
            j++; k--; 
        }
        else
        if ((int)small.size() - i > 0){
            ans = (ans * (small[small.size() - 1] + inf) % inf) % inf;
            //cout << " - " << small[small.size() - 1] << "\n";
            small.pop_back(); k--;
        }
    }
    cout << ans;
    return 0;
}