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 <iostream>
#include <set>
#include <map>
#include <cmath>
#include <vector>
#include <string>
#include <algorithm>
#include <iomanip>
#include <queue>
#include <deque>
#include <stack>
#include <climits>
#include <math.h>
#include <cassert>
#include <unordered_map>
#include <chrono>
#include <random>
#include <ctime>
using namespace std;
#pragma warning(disable : 4996)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

const ld PI = 3.14159265358979323846;
const ll MOD = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll mul(ll a, ll b) {
	return (a * b) % MOD;
}

bool cmp(int a, int b) {
	return abs(a) < abs(b);
}

int main() {
#ifdef _DEBUG
	freopen("input1.txt", "r", stdin);
	freopen("output2.txt", "w", stdout);
#endif
	srand(time(0));
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	vector<ll> a(n);
	int less = 0, eq = 0, more = 0;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		if (a[i] > 0)more++;
		else if (a[i] < 0)less++;
		else eq++;
	}
	sort(a.begin(), a.end());
	ll res = 1, l = 0, r = n - 1;
	bool ok = false;
	for (int i = 0; i <= more; i++) {
		if ((k - i) % 2 == 0 && less >= (k - i))ok = true;
	}
	if (ok) {
		while (k) {
			if (k > 1) {
				if (a[l] * a[l + 1] > a[r] * a[r - 1]) {
					res = mul(res, mul(a[l], a[l + 1]));
					k -= 2;
					l += 2;
				}
				else {
					res = mul(res, a[r]);
					r--;
					k--;
				}
			}
			else {
				res = mul(res, a[r]);
				r--;
				k--;
			}
		}
		res = (res + MOD) % MOD;
		assert(res >= 0);
		cout << res;
	}
	else if (eq > 0) {
		cout << 0;
	}
	else {
		sort(a.begin(), a.end(), cmp);
		for (int i = 0; i < k; i++) {
			res = mul(res, a[i]);
		}
		cout << (res + MOD) % MOD;
	}
}