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 int             long long
#define ld              long double
#define mp              make_pair
#define ff              first
#define ss              second
#define pb              push_back
#define fo(i,n)         for(i=0;i<n;i++)
#define fo1(i,n)        for(i=1;i<=n;i++)
#define FO(i,k,n)       for(i=k;k<=n?i<=n:i>=n;k<=n?i++:i--)
#define all(c)          c.begin(),c.end()
#define rall(c)         c.rbegin(),c.rend()
#define si              set<int>
#define vi              vector<int>
#define mii             map<int,int>
#define pii             pair<int,int>
#define vpii            vector<pii>
#define vvi             vector<vi>
#define usi             unordered_set<int>
#define umii            unordered_map<int,int>
#define pqmx            priority_queue<int>
#define pqmn            priority_queue<int,vi,greater<int> >
#define sp(x)           cout<<setprecision(x)<<fixed;
#define setbits(x)      __builtin_popcountll(x)
#define zerobits(x)     __builtin_ctzll(x)
#define in(a,n)         vi a(n);fo(i,n)cin>>a[i];
#define in1(a,n)        vi a(n+1);fo1(i,n)cin>>a[i];
#define w(t)            int t;cin>>t;while(t--)
#define sz(v)           (int)v.size()
#define gcd             __gcd
#define lcm(x,y)        x*(y/gcd(x,y))
#define endl            '\n'
#define PI              3.14159265358979323846
const int inf =         1e18;
const int mod =         1e9+7;
void __p(int x) {cout<<x;}
void __p(int32_t x){cout<<x;}
void __p(float x){cout<<x;}
void __p(double x){cout<<x;}
void __p(ld x) {cout<<x;}
void __p(char x){cout <<'\''<<x<<'\'';}
void __p(const char *x){cout <<'\"' <<x<<'\"';}
void __p(const string &x) {cout<<'\"'<<x<<'\"';}
void __p(bool x) {cout<<(x?"true":"false");}
void _p() {cout << "]\n";}
template<typename T,typename V>     void __p(const pair<T,V> &x) {cout<<'{';__p(x.first);cout<<',';__p(x.second);cout<<'}';}
template<typename T>                void __p(const T &x) {int f = 0; cout << '{'; for (auto &i: x) cout << (f++ ? "," : ""), __p(i); cout << "}";}
template <typename T,typename... V> void _p(T t, V... v) {__p(t);if(sizeof...(v))cout<<", ";_p(v...);}
int po(int x,int p,int m = inf) {int ans=1;x%=m;while(p>0){if(p&1)ans=(ans*x)%m;p=p>>1;x=(x*x)%m;}return ans;}
int inv(int x,int m=mod){return po(x,m-2,m)%m;}
#ifndef ONLINE_JUDGE
#define dbg(x...) cout<<"["<<#x<<"]=[";_p(x)
#else 
#define dbg(x...)
#endif
//=================================================================================




const int N=1e6+6;
int n,m,k,x,y,z,a,b,c,d,i,j,l,r,f;








void sol()
{   
    cin>>n>>k;
    vi v1,v2;
    a=c=b=0;
    fo(i,n)
    {
        cin>>x;
        if(x>0)v1.pb(x),a++;
        else if(x<0)v2.pb(-x),b++;
        else c++;
    }
    
    if(k>a+b){cout<<0;return;}
    if(c && a+2*(b/2)<k){cout<<0;return;}

    if(c && !a && k&1){cout<<0;return;}
    f=0;
    
    sort(rall(v2));
    if((a+2*(b/2)<k))f=1;
    else if(!a && k&1 ) f=1;
    else if(b&1 && a+b-1>=k)v2.pop_back();
    



    for(auto c:v2)v1.pb(c);
    if(f)sort(all(v1));
    else sort(rall(v1));

    int p=1;
    fo(i,k)
    {
        p=(p*v1[i])%mod;
    }

    if(f)p*=-1;

    p=(p%mod+mod)%mod;

    cout<<p<<endl;

}





int32_t main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //sp(10);
    // w(t)
    sol();
    return 0;
}