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)

//E 
#include <bits/stdc++.h>
#define int long long
using namespace std;
typedef long long ll;
const int maxn = 2e5+10;
const int p = 1e9+7;
ll raw[maxn]; 
struct Node{
	ll w,cnt;
	Node(ll w = 0,ll cnt = 0):w(w),cnt(cnt){}
	bool operator < (Node & x){
		return w == x.w?cnt < x.cnt:w < x.w;
	}
}node[maxn];
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,k;
	cin >> n >> k;
	int postive = 0,neg = 0;
	for(int i = 1;i <= n;i++){
		cin >> raw[i];
		if(raw[i] > 0)postive++;
		else if(raw[i] < 0)neg++;
	}
//	cout << postive << endl;
	if(!postive){
		sort(raw+1,raw+n+1);
		ll ans = 1;
		if(neg%2)for(int i = 1;i <= k;i++)ans = (ans*raw[i]%p + p)%p;
		else for(int i = n;i >= n-k+1;i--)ans = (ans*raw[i]%p + p)%p;
		cout << ans;
		return 0;
	}
	int qwq = 0;

	sort(raw+1,raw+n+1);
	for(int i = 1;i <= n;i++)if(raw[i] >= 0){
		qwq = i;break;
	}
	reverse(raw+1,raw+qwq);
	int idx = 0,i = 1;
	if(neg%2)i = 2;
	ll mul = 0;
	for(;i <= n;i++){
		if(raw[i] < 0){
			if(!mul)mul = raw[i];
			else{
				 node[++idx] = Node(mul*raw[i],2);
//				 cout << mul*raw[i] << " " << i << " " << mul << " " << raw[i] << endl; 
				mul = 0;
			}
		}
		else node[++idx] = Node(raw[i]*2,1);
	}
//	cout << idx << endl;
	sort(node+1,node+idx+1);
	ll ans = 1;
	while(k){
		if(k >= node[idx].cnt){
			k -= node[idx].cnt;
			int w = node[idx].w;
			if(node[idx].cnt == 1)w /= 2;
//			cout << w << endl;
			ans = (ans*w%p + p)%p;
		}
		idx--;
	}
	cout << (ans%p + p)%p;
	return 0;
}