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: Runtime Error)

#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define mk make_pair
#define all(a) a.begin(), a.end()
#define len(a) (int)a.size()

using namespace std;

typedef long long ll;
typedef vector <int> vi;
typedef pair <int, int> pii;

int MOD = 1e9 + 7;

ll mod(ll a, ll b){ 
	return (a % b + b) % b; 
}

int main(){
	int n, k;
	cin >> n >> k;
	
	vector <int> d(n);
	for(int i = 0; i < n; i++)
		cin >> d[i];
		
	vector <ll> p, m, mm;
	
	int zero = 0;
	
	for(int i = 0; i < n; i++)
		if(d[i] < 0){
			m.pb(d[i]);
			mm.pb(d[i]);
		}
		else if(d[i] > 0)
			p.pb(d[i]);
		else
			zero++;
		
	sort(all(p));
	reverse(all(p));
	sort(all(m));
	sort(all(mm));
	reverse(all(mm));
	
	for(int i = 1; i < len(p); i++)
		p[i] = (p[i] * p[i-1]) % MOD;
	for(int i = 1; i < len(m); i++)
		m[i] = (m[i] * m[i-1]) % MOD;
	for(int i = 1; i < len(mm); i++)
		mm[i] = (mm[i] * mm[i-1]) % MOD;
		
	if(k > len(p) + len(m)){
		cout << 0 << endl;
		return 0;
	}
	
	p.insert(p.begin(), 1);
	m.insert(m.begin(), 1);
	mm.insert(mm.begin(), 1);
	
	ll mx = (zero > 0 ? 0 : -1e9);
	for(int i = 0; i < len(p); i++){
		int idx = k - i;
		if(idx > len(m))
			continue;
		if(idx % 2 == 0)
			mx = max(mx, (p[i] * m[idx]) % MOD);
		else{
			mx = max(mx, (p[i] * mm[idx]) % MOD);
		}
	}
	
	cout << mod(mx, MOD) << endl;
	return 0;
}