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;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = (1<<30) - 1;
const ll LINF = (1LL<<60) - 1;
const int mod = 1e9 + 7;

int main(){
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    sort(a.begin(), a.end(), [](int a, int b) {return abs(a) > abs(b); });

    ll res = 1;
    int sign = 1;
    bool eqzero = false;
    for (int i = 0; i < k; ++i) {
        res *= a[i];
        if (a[i] < 0) sign *= -1;
        if (a[i] == 0) eqzero = true;
        res %= mod;
    }

    if (sign > 0 || eqzero){
        res += mod;
        res %= mod;
        cout << res << endl;
        return 0;
    }

    res = 1;
    ll lpos = INF;
    ll lneg = INF;
    ll rpos = INF;
    ll rneg = INF;
    for (int i = k - 1; i >= 0; --i) {
         if (a[i] > 0){
             lpos = a[i];
             break;
         }
    }
    for (int i = k - 1; i >= 0; --i) {
        if (a[i] < 0){
            lneg = a[i];
            break;
        }
    }
    for (int i = k; i < n; ++i) {
        if (a[i] > 0){
            rpos = a[i];
            break;
        }
    }
    for (int i = k; i < n; ++i) {
        if (a[i] < 0){
            rneg = a[i];
            break;
        }
    }

    if (lpos == INF && rpos == INF){
        reverse(a.begin(), a.end());
        for (int i = 0; i < k; ++i) {
            res *= a[i];
            res %= mod;
        }
        res += mod;
        res %= mod;
        cout << res << endl;
        return 0;
    }

    int out, in;
    if (lpos != INF && rpos != INF && lneg != INF && rneg != INF){
        if (lneg * rneg > lpos * rpos){
            out = lpos;
            in = rneg;
        } else{
            out = lneg;
            in = rpos;
        }
    } else if (lneg != INF && rpos != INF){
        out = lneg;
        in = rpos;
    } else{
        out = lpos;
        in = rneg;
    }

    bool flag = true;
    for (int i = 0; i < k; ++i) {
        if (a[i] == out && flag){
            res *= in;
            res %= mod;
            flag = false;
            continue;
        }
        res *= a[i];
        res %= mod;
    }

    res += mod;
    res %= mod;
    cout << res << endl;
}