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 ll long long
#define lvector vector<ll>
#define cvector vector<char>
#define svector vector<string>
#define lque queue<ll>
#define lpque priority_queue<ll>
#define dlpque priority_queue<ll,lvector,greater<ll>>
#define P pair<ll,ll>
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define rep(i,n) for(ll i=0; i<n; ++i)
#define print(a) cout << (a) << endl

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n,k,prod=1,mod=1e9+7,ps,ns,x;cin>>n>>k;
  lvector A(n),pos,neg;
  rep(i,n) {
    cin>>A[i];
    if(A[i]<0) neg.emplace_back(A[i]);
    else pos.emplace_back(A[i]);
  }
  ps=pos.size(),ns=neg.size();
  bool ispos=false;
  if(ps>0) {
    if(n==k) ispos=(ns%2==0);
    else ispos=true;
  } else ispos=(k%2==0);
  if(!ispos) {
    sort(ALL(A),[](ll x,ll y) {return abs(x)<abs(y);});
    rep(i,k) {
      prod*=A[i];
      while(prod<0) prod+=mod;
      prod%=mod;
    }
  }
  else {
    sort(ALL(pos));sort(RALL(neg));
    if(k&1) {
      prod*=pos.back();
      prod=(prod+mod)%mod;
      pos.pop_back();
    }
    lvector V;
    while(pos.size()>=2) {
      x=pos.back();pos.pop_back();
      x=x*(pos.back())%mod;pos.pop_back();
      V.emplace_back(x);
    }
    while(neg.size()>=2) {
      x=neg.back();neg.pop_back();
      x=x*(neg.back())%mod;neg.pop_back();
      V.emplace_back(x);
    }
    sort(RALL(V));
    rep(i,k/2) {
      prod*=V[i];
      prod%=mod;
    }
  }
  print(prod);
  return 0;
}