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: Accepted)

#include <bits/stdc++.h>
#define int long long
using namespace std;

int mod = 1e9 + 7;

signed main(){
  
  int n, k, zero = 0;
  vector<int> A, B;
  
  cin >> n >> k;
  
  for(int i=0;i<n;i++){
    int a;
    cin >> a;
    if( a == 0 ) zero++;
    if( 0 < a ) A.push_back( a );
    if( a < 0 ) B.push_back( -a );
  }
  
  sort( A.begin(), A.end(), greater<int>() );
  sort( B.begin(), B.end(), greater<int>() );
  
  // pattern 1
  if( n - k < zero ){
    cout << 0 << endl;
    return 0;
  }
  
  // pattern 2
  int mul = 1, cnt = k;
  int idxA = 0, idxB = 0;
  
  while( cnt ){

    int Acnt = (int)A.size() - idxA;      
    int Bcnt = (int)B.size() - idxB;
    
    if( cnt % 2 ){
      
      if( Acnt ){
	mul = mul * A[idxA++] % mod;
	cnt--;
      }
      else break;
      
    }
    else{
      
      if( 2 <= Acnt && 2 <= Bcnt ){
	
	if( A[idxA] * A[idxA+1] < B[idxB] * B[idxB+1] ){
	  mul = mul * B[idxB++] % mod;
	  mul = mul * B[idxB++] % mod;
	  cnt -= 2;
	}
	else{
	  mul = mul * A[idxA++] % mod;
	  mul = mul * A[idxA++] % mod;
	  cnt -= 2;
	}
	
      }
      else if( 2 <= Acnt ){
	mul = mul * A[idxA++] % mod;
	mul = mul * A[idxA++] % mod;
	cnt -= 2;
      }
      else if( 2 <= Bcnt ){
	mul = mul * B[idxB++] % mod;
	mul = mul * B[idxB++] % mod;
	cnt -= 2;
      }
      else break;
    }
    
  }
  
  if( cnt == 0 ){
    cout << mul << endl;
    return 0;
  }
  else if( zero ){
    cout << 0 << endl;
    return 0;
  }
  
  // pattern 3

  sort( B.begin(), B.end() );
  
  if( n == k ){
    
    mul = 1;
    
    for(int a : A ) mul = mul * a % mod;
    for(int b : B ) mul = mul * b % mod;
    
    cout << mod - mul << endl;
    
  }
  else{
    
    mul = 1;

    for(int i=0;i<k;i++) mul = mul * B[i] % mod;
    
    cout << mod - mul << endl;
      
  }
  
  return 0;
}