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;

typedef long long ll;
typedef pair<ll,ll> pi;
typedef vector <ll> vi;
typedef vector <pi> vpi;
#define f first
#define s second
#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)
#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0)
#define maxn 200001
#define MOD 1000000007
#define int ll
int N,K;
int A[maxn],B[maxn];
double ass[maxn],bss[maxn];

int32_t main(){
	fast;
	
	cin>>N>>K;
	int posidx = 1, negidx = 1;
	FOR(i,1,N){
		int a; cin>>a;
		if (a > 0){
			A[posidx] = a;
			posidx++;
		}else{
			B[negidx] = a;
			negidx++;
		}
	}
	posidx--, negidx--;
	sort(A+1,A+posidx+1, greater<int>());
	sort(B+1,B+negidx+1,greater<int>());
	
	
	if (posidx == 0){
		int ans = 1;
		if (K % 2 == 0) sort(B+1,B+negidx+1);
		
		FOR(i,1,K){
			ans *= B[i];
			ans %= MOD;
		}
		
		
		cout<<(ans + MOD) % MOD;
		return 0;
	}

	FOR(i,1,posidx) ass[i] = ass[i-1] + log10(A[i]);
	FOR(i,1,negidx) bss[i] = bss[i-1] + log10(-B[i]);
		
	double ans = 0; int legans = 0;
	for (int i = 0; i <= K; i += 2){
		if (i > negidx || K - i > posidx) continue;
		if (ass[K - i] + bss[i] > ans){
			ans = max(ans, ass[K - i] + bss[i]);
			legans = i;
		}
	}

	int actlans = 1;
	
	FOR(i,1,legans){
		actlans *= B[i];
		actlans %= MOD;
	}
	FOR(i,1,K - legans){
		actlans *= A[i];
		actlans %= MOD;
	}
	
	cout<<actlans;
}