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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <tuple>
#include <cstdio>
#include <cmath>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;

int mod = 1000000007;
int n, k;
int a[200000];
vector<int> sei; //正の値の絶対値
vector<int> fu;	 //負の値の絶対値

signed main() {
	int i;
	
	cin >> n >> k;
	rep(i, n) {
		cin >> a[i];
		if (a[i] >= 0) sei.push_back(a[i]);
		else fu.push_back(-a[i]);
	}
	
	//答えが負になるケース
	if (sei.size() == 0 && k % 2 == 1) {
		sort(fu.begin(), fu.end());
		int ans = 1;
		rep(i, k) {
			ans *= fu[i];
			ans %= mod;
		}
		cout << (mod - ans) % mod << endl;
		return 0;
	}
	
	//答えが正（負からは偶数個取る）
	//kが奇数→seiから最大値を1個取る
	//kが偶数→どちらからも偶数個取る. 2個ずつ積でまとめ、大きい方から貪欲.
	sort(sei.begin(), sei.end(), greater<int>());
	sort(fu.begin(), fu.end(), greater<int>());
	
	int ans = 1;
	if (k % 2 == 1) {
		ans *= sei[0];
	}
	
	vector<int> vec;
	for (i = k % 2; i + 1 < sei.size(); i += 2) {
		vec.push_back(sei[i] * sei[i + 1]);
	}
	for (i = 0; i + 1 < fu.size(); i += 2) {
		vec.push_back(fu[i] * fu[i + 1]);
	}
	sort(vec.begin(), vec.end(), greater<int>());
	
	rep(i, k / 2) {
		int res = vec[i] % mod;
		ans *= res;
		ans %= mod;
	}
	cout << ans << endl;
	return 0;
}