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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(ll i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second
constexpr char ln = '\n';

const int mx=200010;
const ll mod=1e9+7;

struct mint {
    ll x; // typedef long long ll;
    mint(ll x=0):x((x%mod+mod)%mod){}
    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 { mint res(*this); return res+=a; }
    mint operator-(const mint a) const { mint res(*this); return res-=a; }
    mint operator*(const mint a) const { mint res(*this); return res*=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 { mint res(*this); return res/=a; }
};

int main(){
  int n,k;
  cin >> n >> k;
  vector<pair<ll,int>> a(n); 
  mint sum = 1;
  bool flag = false;
  rep(i,n){
    cin >> a[i].F;
    sum *= a[i].F;
    if(a[i].F>=0){
      a[i].S = 0;
      flag = true;
    } else {
      a[i].F *= -1;
      a[i].S = 1;
    }
  }
  if(k==n){
    cout << sum.x << ln; return 0;
  }
  if(k%2 && !flag){
    sort(a.begin(),a.end());
    mint ans = 1;
    rep(i,k){
      ans *= (-1)*a[i].F;
    }
    cout << ans.x << ln; return 0;
  }

  sort(a.rbegin(),a.rend());
  mint ans = 1;
  int cnt = 0;
  rep(i,k){
    ans *= a[i].F;
    cnt += a[i].S;
  }

  if(cnt%2==0 || ans.x==0){
    cout << ans.x << ln; return 0;
  }
  ll A=inf,B=inf,C=inf,D=inf;
  REP(i,k,n){ if(a[i].S==0){ C = a[i].F; break; } }
  REP(i,k,n){ if(a[i].S==1){ D = a[i].F; break; } }
  for(int i=k-1; i>=0; i--){ if(a[i].S==0){ B = a[i].F; break; } }
  for(int i=k-1; i>=0; i--){ if(a[i].S==1){ A = a[i].F; break; } }
  if((C==inf || A== inf)&& (B!=inf && D!= inf)){
    ans *= D;
    ans /= B;
  }
  if((B==inf || D== inf) && (A!=inf && C!= inf)){
    ans *= C;
    ans /= A;
  }
  if(B!=inf && D!= inf && A!=inf && C!= inf){
    if(C*D >= A*B ){
      ans *= C;
      ans /= A;
    }
    if(C*D < A*B ){
      ans *= D;
      ans /= B;
    }
  }
  cout << ans.x << ln;
  return 0;
}