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>
#define rep(i,n) for (int i = 0; i < (n); i++)
#define rng(a) a.begin(),a.end()
using namespace std;
using ll = long long;
const int INF = 1001001001;
const int mod = 1000000007;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
  return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
  mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }


  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

int main() {
  int n, K;
  cin >> n >> K;
  int k = K;
  vector<int> p, m;
  rep(i,n) {
    int x; cin >> x;
    if(x>=0) p.push_back(x);
    else m.push_back(-1*x);
  }
  sort(p.rbegin(),p.rend());
  sort(m.rbegin(),m.rend());
  int pp = 0, mm = 0;
  mint ans = 1;
  int s = p.size(), t = m.size();
  if(k%2==1&&s>0) {
    ans *= p[pp]; pp++;
  }
  while(k>0&&pp<s-1&&mm<t-1) {
    ll P = p[pp]*p[pp+1], M = m[mm]*m[mm+1];
    if(P<M) {
      ans *= M; mm += 2;
    }
    else {
      ans *= P; pp += 2;
    }
    k -= 2;
  }
  if(s+t==k) {
    while(k>0) {
      if(pp<s) {
        ans *= p[pp];  pp++;
      }
      else if(mm<t) {
        ans *= m[mm];  mm++;
      }
      k--;
    }
  }
  //正の数が0個の時を除き、kは偶数
  else if(s==0&&k%2==1) {//ans<0
    ans = 1;
    reverse(rng(m));
    rep(i,K) {
      ans *= -1*m[i];
    }
  }
  else if(pp>=s-2) { //正の数は0または1個
    while(k>0) {
      ans *= m[mm];
      mm++; k--;
    }
  }
  else if(mm>=s-2) {
    while(k>0) {
      ans *= p[pp];
      pp++; k--;
    }
  }
  
  cout << ans << endl;
  return 0;
}