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)

/**
 *    author:   K.R.Surrya  
 *    created:  05.07.2020 18:06:16
**/
//~ ~~ ~~~~ ~~~~~~~~   
#include<bits/stdc++.h>
using namespace std; 
using lint  =                           long long int;
using pint  =                           pair<int, int>;
using plint =                           pair<lint, lint>;
#define arr                             array
#define vi                              vector<int>
#define vli                             vector<lint>
#define pb                              push_back
#define ALL(x)                          begin(x),end(x)
#define inf                             (1LL<63)-1
#define MOD                             1000000007
#define SZ(x)                           ((lint)(x).size())
#define                                 FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define                                 IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define                                 REP(i, n) FOR(i,0,n)
#define                                 IREP(i, n) IFOR(i,0,n)
#define                                 f_ip   ios_base::sync_with_stdio(false); cin.tie(NULL);
#define ZERO(a)                         fill(ALL(a),0)
#define lastBit(n)                      31-__builtin_clz(n) + 1 //32 bit integers, 1-based indexing.
#define firstBit(n)                     __builtin_ctz((n&(-n))) + 1
struct fast_ios                         { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
template<typename T>                    istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T>                    bool chmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T>                    bool chmin(T &m, const T q) { if (q < m) {m = q; return true;} else return false; }
mt19937                                 rng(chrono::steady_clock::now().time_since_epoch().count());



        /* ~Debugging~ */         
      /*~~~~~~~~~~~~~~~~~*/       
#define LOCAL
#define sim template <class c
#define ris return *this
#define dor > deb &operator<<
#define eni(x)                                                                  \
    sim > typename enable_if<sizeof dud<c>(0) x 1, deb &>::type operator<<(c i) \
    {
sim > struct rge
{
    c b, e;
};
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c *x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct deb
{
#ifdef LOCAL
    ~deb()
    {
        cerr << endl;
    }
    eni(!=) cerr << boolalpha << i;
    ris;
} eni(==) ris << range(begin(i), end(i));
}
sim, class b dor(pair<b, c> d)
{
    ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d)
{
    *this << "[";
    for (auto it = d.b; it != d.e; ++it)
        *this << ", " + 2 * (it == d.b) << *it;
    ris << "]";
}
#else
    sim dor(const c &)
    {
        ris;
    }
#endif
}
;
#define what(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "
// -DLOCAL



      /*  ~magic~ */            
    /*~~~~~~~~~~~~~~~*/           
#define magic(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
void err(istream_iterator<string> it)  { return ;}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a,Args... args)
{
cerr << *it << " = " << a <<endl;
err(++it,args...);   //args is going to end.
}


    /* ~driver~ */
    /*~~~~~~~~~~*/

const int mod = (int)1e9 + 7;
const int INF = INT_MIN;
lint mul(lint& a,lint& b) { 
    lint ret = (a%mod*b%mod)%mod;
    /*
    if(ret<0) { 
        ret += mod;
    }*/
    return ret;
}

lint getMod(lint& ret) { 
    ret = ret%mod;
    if(ret<0) ret+=mod;
    return ret;
}

void solve() {
     lint n;cin >> n; 
     lint k;cin >> k;
     vli a(n);
     vector<lint> pos;
     vector<lint> neg;
     int zero = 0;  //out of the game rn;;;
      
     REP(i,n) { 
         cin >> a[i]; //taking input into the array;;;

         if(a[i]==0) { 
             ++zero;
            pos.push_back(a[i]);
         } else if(a[i]>0) { 
             pos.push_back(a[i]);
         } else {
             neg.push_back(a[i]);
         }
     }

     sort(pos.rbegin(),pos.rend()); // descending order ;;;
     sort(ALL(neg));                // ascending order ;;; ( descending in magnitude ).

     vector<lint> ap((int)pos.size()+1);
     vector<lint> an((int)neg.size()+1);

     ap[0]=(lint)1;
     an[0]=(lint)1;

     REP(i,(int)pos.size()) { 
         ap[i+1]=mul(ap[i],pos[i]);
     }
     REP(i,(int)neg.size()) { 
         an[i+1]=mul(an[i],neg[i]);
     }

     lint ret = INF;
     //prevention : 
    //deb() << what(k) << what((int)pos.size()) << what((int)neg.size()) << what(zero);
     if((int)pos.size()==0) { 
       if((k&1)) { 
                 lint cc = 0; 
                 lint answer = (lint)1;
                 IREP(i,(int)neg.size()) { 
                      if(cc<k) { 
                        //deb() << what(neg[i]);    //smaller value in negative ==> bigger number;;;
                         answer = mul(answer,neg[i]);
                         ++cc;
                      } else { 
                          break;
                      }
                 }
                 answer = getMod(answer);
                 cout << answer << '\n';
                 return;
       }  
     }

    //  deb() << what(1);

     REP(i,k+1) { 
         if(i<(int)an.size()) { 
             lint one = an[i];
             if(k-i < (int)ap.size()) { 
                 lint two = ap[k-i];
                 lint temp = mul(one,two);
                //  deb() << what(i) << what(one) << what(k-i) << what(one) << what(two) << what(temp);
                 chmax(ret,temp);
             }
         }
     }

    //  if(ret == INF) { 
    //      assert(!zero); //maybe this is causing the run-time error;;;
    //      cout << 0 << '\n';
    //      return ;
    //  }
    //  if(ret<0 && (zero>0)) {  // test-case :  { 1,2,3,-1,-2,-3  and 0} k = 6, this is also working fine ;;; 
    //      cout << 0 << '\n';
    //      return;
    //  }
    // deb() << what(ret);

    //minor change for negative result;;
    //smaller values with smaller values positive values;;;
    //-ve implies that there must not be any zeros involved;

    if(ret<0) {
        if(zero) { 
            cout << 0 << '\n';
            return;
        }
    }

    //now try to maximize the negative-result;;;
    reverse(ALL(neg));
    reverse(ALL(pos));
    REP(i,k+1) { 
         if(i<(int)an.size()) { 
             lint one = an[i];
             if(k-i < (int)ap.size()) { 
                 lint two = ap[k-i];
                 lint temp = mul(one,two);
                //  deb() << what(i) << what(one) << what(k-i) << what(one) << what(two) << what(temp);
                 chmax(ret,temp);
             }
         }
     }

    ret = getMod(ret);
    cout << ret;
}

int main()
{   
     #if 0
     freopen("input.txt", "r", stdin);
     freopen("output.txt", "w", stdout);
     #endif

     int T=1;
     //cin>>T;
     for(int tt=1;tt<=T;++tt) {
         solve();
     }

    return 0;
}


     /* ~Modulus functions~ */
    /*~~~~~~~~~~~~~~~~~~~~~~~*/
lint pow_mod(lint base,lint expo) { 
   lint ret = 1;
   for(;expo;) { 
      if(expo&1)  ret = (ret*base)%MOD; 

      expo = (expo>>1);
      base = (base*base)%MOD;
   }
   return ret;
}