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>
#define all(x) (x).begin(), (x).end()
typedef long long ll;
#define MOD 1000000007
using namespace std;
ll sign(ll n) {
    if(n > 0) {
        return 1;
    }
    if(n == 0) {
        return 0;
    }
    return -1;
}
int main() {
    ll n, k;
    cin >> n >> k;
    vector<pair<ll, ll>> a(n);
    for(int i = 0; i < n; i++) {
        ll b;
        cin >> b;
        a[i].first = abs(b);
        a[i].second = sign(b);
    }
    sort(all(a));
    reverse(all(a));
    ll ans = 1;
    ll m = 0;
    for(int i = 0; i < n; i++) {
        ans *= a[i].first;
        ans %= MOD;
        if(a[i].second < 0) {
            m++;
        }
        if(m % 2 == 0 && i + 1 == k) {
            cout << ans << endl;
            return 0;
        }
    }
    ll cnt = 0;
    ans = 1;
    for(int i = 0; i < n; i++) {
        if(a[i].second > 0) {
            ans *= a[i].first;
            ans %= MOD;

            cnt++;
        }
        if(cnt == k) {
            cout << ans << endl;
            return 0;
        }
    }
    if(a.back().first == 0) {
        return 0;
    } else {
        ans = 1;
        for(int i = k - 1; i >= 0; i--) {
            ans *= a[i].first;
            ans %= MOD;
        }
        cout << ans << endl;
        return 0;
    }
}