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)

/* winners never quit and quitters never win.
      #swap                                 */
      
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp> 
using namespace std;
using namespace __gnu_pbds;
#pragma GCC optimize ("-O3")
using ll = long long;
using ld = long double;
const ll mod = 1000000007;
const ll inf = 1000000000000000000;
const ll rk = 256;
const ld PI = 3.141592653589793;

typedef tree<
int,
null_type,
less<int>,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;

template <typename T>
void make_unique(vector<T>& vec) {
  sort(all(vec));
  vec.erase(unique(all(vec)), vec.end());
}

#define pb push_back
#define mp make_pair
#define vc vector
#define fs first
#define sec second
#define pq priority_queue
#define lb lower_bound
#define ub upper_bound
#define pll pair<ll,ll> 
#define pls pair<ll,string>
#define psl pair<string,ll>
#define plc pair<ll,char>
#define pcl pair<char,ll>
#define pss pair<string,string>
#define all(x) (x).begin(), (x).end()
#define tol(s) transform(s.begin(),s.end(),s.begin(),::tolower);
#define tou(s) transform(s.begin(),s.end(),s.begin(),::toupper);
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";

#define error(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) {}
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " = " << a << endl;
	err(++it, args...);
}

#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define T ll t=0;cin>>t;for(ll test=0;test<t;test++)

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll nxt() {
    ll TemporaryVariable;
    cin >> TemporaryVariable;
    return TemporaryVariable;
}

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

ll fast_exp(ll base, ll exp1) {
    base%=mod;
    ll res=1;
    while(exp1>0) {
       if(exp1&1) res=(res*base)%mod;
       base=(base*base)%mod;
       exp1/=2;
    }
    return res%mod;
} 

ll add(ll x,ll y)
{
    return (x+y+mod)%mod;
}

ll mul(ll x,ll y)
{
    while(x<0)
        x+=mod;
    while(y<0)
        y+=mod;
    return (x*y)%mod;
}

 
int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
	//isprime();
	//coeff();
	/*freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);*/
    ll n=nxt(),k=nxt();
    vc<ll> v(n);
    generate(all(v),nxt);
    sort(all(v));
    ll ans=1,last=n-1;
    if(k%2==1)
    {
        if(v[n-1]<=0)
        {
            ll p=n-1;
            while(k)
            {
                ans=mul(ans,v[p]);
                k--;
                p--;
            }
            cout<<ans<<endl;
            exit(0);
        }
        ans=mul(ans,v[n-1]);
        last--;
        k--;
    }
    ll start=0;
    while(k)
    {
        ll p=v[last]*v[last-1],q=v[start]*v[start+1];
        if(q>p)
        {
            ans=mul(ans,mul(q,1LL));
            start+=2;
        }
        else
        {
            ans=mul(ans,mul(p,1LL));
            last-=2;
        }
        k-=2;
    }
    cout<<ans<<endl;
	printclock;
	return 0;
}