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: Accepted)

#include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;

#define int long long int
#define double long double
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define rep(i,a,b) for (int i = a; i < b; ++i)
#define IOS ios::sync_with_stdio(false); cin.tie(NULL);
#ifndef LOCAL
#define dbg(...) ;
#define endl '\n'
#endif

const int inf = 1e15;
const int MOD = 1e9 + 7;
const int N = 2e5 + 5;



signed main(){
    IOS;
    int n, k;
    cin >> n >> k;
    int a[n];
    for (int i = 0; i < n; i++)
        cin >> a[i];
    sort(a, a + n);
    int ans = 1;
    if (k % 2 == 1) {
        if (a[n-1] == 0) return cout << 0 << endl, 0;
        else if (a[n-1] < 0) {
            int p = 1;
            for (int i = n - 1; i >= n - k; i--) {
                p = (p * a[i] % MOD + 10 * MOD) % MOD;
            }
            cout << p << endl;
            return 0;
        }
        else {
            k--;
            ans = ans * a[n-1] % MOD;
            n--;
        }
    }
    if (k % 2 == 0) {
        int l = 0, r = n - 1;
        for (int i = 0; i < k / 2; i++) {
            if (a[l] * a[l+1] > a[r] * a[r-1]) {
                ans = ans * (a[l] * a[l+1] % MOD + 10 * MOD) % MOD;
                l += 2;
            }
            else {
                ans = ans * (a[r] * a[r-1] % MOD + 10 * MOD) % MOD;
                r -= 2;
            }
        }
    }
    cout << ans << endl;
    return 0;
}