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 <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define __ ios_base::sync_with_stdio(0);cin.tie(0);
#define fi first
#define se second
#define pb push_back
#define all(x) x.begin(),x.end()
#define forn(i,a,n) for(int i=a; i < n; i++)
typedef long long int lli;
typedef long double Double;
typedef pair<lli,lli> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
const int mod = 1e9+7;
const lli inf = 2e18;
int main(){__
	int n,k;
	cin >> n >> k;
	vector<lli> neg,pos;
	lli acum = 1ll;
	for(int i=0; i < n; i++){
		lli x;
		cin >> x;
		acum = (acum * x) % mod;
		acum%=mod;
		if(acum <0) acum+=mod;

		if(x < 0) neg.pb(x);
		else pos.pb(x);
	}
	
	if(k == n){
		cout << acum << endl;
		return 0;
	}
	
	if(!neg.empty()) sort(all(neg));
	if(!pos.empty()) sort(all(pos));
	reverse(all(neg));
	if(pos.empty()){
		if(k%2 == 1){
			lli res = 1ll;
			for(int i=0,c=0; i<n && c < k;i++,c++){
				res = (res * neg[i]) % mod;
				res%=mod;
				if(res < 0) res+=mod;
				res%=mod;
			}
			res%=mod;
			if(res < 0) res+=mod;
			cout << res << endl;
		}
		else{
			lli res = 1ll;
			for(int i=n-1,c=0; i>=0 && c < k;i--,c++){
				res = (res * neg[i]) % mod;
				res%=mod;
				if(res < 0)res+=mod;
				res%=mod;
			}
			cout << res << endl;
		}
		return 0;
	}
	lli ans = 1ll;
	int szn = (int)neg.size(), szp = (int)pos.size();
	while(k > 1){
		lli prodneg = 1ll,prodpos =1ll;
		if(szn){
			prodneg *= neg[szn-1];
			if(szn > 1) prodneg*= neg[szn-2];
		}
		else{
			prodneg = -inf;
		}

		if(szp){
			prodpos*= pos[szp-1];
			if(szp>1) prodpos*=pos[szp-2];
		}
		else prodpos = -inf;
		
		if(prodneg > prodpos){
			ans = (ans * prodneg) % mod;
			ans%=mod;
			if(ans < 0) ans+=mod;
			ans%=mod;
			neg.pop_back();
			neg.pop_back();
			szn-=2;
			k-=2;
		}
		else{
			ans = (ans * pos[szp-1]) % mod;
			ans%=mod;
			if(ans < 0) ans+=mod;
			ans%=mod;
			pos.pop_back();
			szp--;
			k--;
		}
	}
	if(k){
		assert(szp > 0);
		ans = (ans * pos[szp-1]) % mod;
	}
	ans%=mod;
	if(ans < 0) ans+=mod;
	ans%=mod;
	cout << ans%mod << endl;
	return 0;
}