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 FOR(i, a, b) for(auto i=a; i<=b; ++i)
#define REP(i, a, b) for(auto i=a; i<b; ++i)
#define FORI(i, a, b) for(auto i=a; i!=b+1-2*(a>b); i+=1-2*(a>b))
#define REPI(i, a, b) for(auto i=a-(a>b); i!=b-(a>b); i+=1-2*(a>b))
#define ALL(v) v.begin(),v.end()
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)
#define pf(a) push_front(a)
#define eb(a, b) emplace_back(a, b)
#define fir first
#define sec second
#define what_is(x) cout<<#x<<" is "<<x<<endl;
#define type(x) typeid(x).name()
#define ms(arr, val) memset(arr, val, sizeof(arr))
#define min3(a,b,c) min(min(a,b),c)
#define max3(a,b,c) max(max(a,b),c)
#define PI acos(-1)
#define open_read freopen("christmas.in", "r", stdin)
#define open_write freopen("christmas.out", "w", stdout)
 
using namespace std;
 
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<double, double> PDD;
typedef pair<int, int> PII;
typedef pair<int, PII> PIPII;
typedef pair<PII, PII> PPIIPII;
typedef pair<LL, LL> PLL;
typedef pair<ULL, ULL> PUU;
typedef pair<LL, PLL> PLPLL;
typedef pair<PLL, PLL> PPLLPLL;
typedef pair<int, LL> PIL;
typedef pair<LL, int> PLI;
 
template<typename T, typename T1>
ostream& operator << (ostream &out, pair<T, T1> obj) {
    out << "(" << obj.first << ", " << obj.second << ")";
    return out;
}
 
template<typename T, typename T1>
ostream& operator << (ostream &out, map<T, T1> cont) {
    typename map<T, T1>::const_iterator itr = cont.begin();
    typename map<T, T1>::const_iterator ends = cont.end();
 
    for(; itr != ends; ++itr) {
        out << *itr << " ";
    }
 
    out<<endl;
    return out;
} 

template<typename T>
ostream& operator << (ostream &out, set<T> cont) {
    typename set<T>::const_iterator itr = cont.begin();
    typename set<T>::const_iterator ends = cont.end();
    
    for(; itr != ends; ++itr) {
        out << *itr << " ";
    }
 
    out << endl;
    return out;
}
  
template<typename T>
ostream& operator << (ostream &out, multiset<T> cont) {
    typename multiset<T>::const_iterator itr = cont.begin();
    typename multiset<T>::const_iterator ends = cont.end();
 
    for (; itr != ends; ++itr) {
        out << *itr << " ";
    }
 
    out << endl;
    return out;
}

template<typename T, template<typename ELEM, typename ALLOC=allocator<ELEM>> class CONT>
ostream& operator << (ostream &out, CONT<T> cont) {
    typename CONT<T>::const_iterator itr = cont.begin();
    typename CONT<T>::const_iterator ends = cont.end();
 
    for (; itr != ends; ++itr) {
        out << *itr << " ";
    }
 
    out << endl;
    return out;
}
 
template<typename T, unsigned int N, typename CTy, typename CTr>
typename enable_if<!is_same<T, char>::value, basic_ostream<CTy, CTr> &>::type
operator << (basic_ostream<CTy, CTr> &out, const T (&arr)[N]) {
    REP (i, 0, N) {
        out << arr[i] << " ";
    }
 
    out << endl;
    return out;
}
  
template<typename T>
T GCD(T a, T b) {
    T min_v = min(a, b);
    T max_v = max(a, b);
 
    while (min_v) {
        T temp = max_v % min_v;
        max_v = min_v;
        min_v = temp;
    }
 
    return max_v;
}
 
template<typename T>
T LCM(T a, T b) {
    return (a * b) / gcd(a, b);
}
 
template<typename T>
T fastExpPow(T base, T exp, T mod) {
    T res = 1;
 
    while (exp) {
        if (exp & 1) {
            res *= base;
            res %= mod;
        }
 
        exp >>= 1;
        base *= base;
        base %= mod;
    }
 
    return res % mod;
}
 
/*#################################################################################################################
###################################################################################################################
###################################################################################################################
#################################################################################################################*/

#define SIZE 200010
#define MOD 1000000007LL 

vector<PLI> numbers;
LL answer;
int cntPos, cntNeg;


int main() {
    int N, K;
    scanf("%d%d", &N, &K);
    
    
    REP (i, 0, N) {
        LL val;
        scanf("%lld", &val);

        if (val > 0) {
            numbers.eb(val, 1);
        } else {
            numbers.eb(-val, -1);
        }

        cntNeg += (numbers[i].sec == -1);
    }

    sort(ALL(numbers), greater<PLI>());

    if (cntNeg == N) {
        answer = 1;

        REP (i, 0, K) {
            answer = (answer * numbers[N - i - 1].fir) % MOD;
            answer = MOD - answer;
        }
        
        printf("%lld\n", answer);
        return 0;
    }

    cntPos = 0, cntNeg = 0;
    answer = 1;
    LL minimNeg = MOD, minimPos = MOD;
    
    REP (i, 0, K) {
        answer = (answer * numbers[i].fir) % MOD;

        if (numbers[i].sec == -1) {
            answer = MOD - answer;
            minimNeg = numbers[i].fir;
            ++cntNeg;
        } else if (numbers[i].sec == 1) {
            minimPos = numbers[i].fir;
            ++cntPos;
        }
    } 
    
    if (cntNeg % 2 == 0) {
        printf("%lld\n", answer);
        return 0;
    }
    
    double change1 = -1, change2 = -1;
    LL answer1 = answer, answer2 = answer;
 
    REP (i, K, N) {
        if (numbers[i].sec == -1 && minimPos != MOD && change1 == -1) {
            change1 = (double)numbers[i].fir / (double)minimPos; 
            answer1 = (answer1 * fastExpPow(minimPos, MOD - 2LL, MOD)) % MOD;
            answer1 = (answer1 * numbers[i].fir) % MOD;
            answer1 = MOD - answer1;
        } else if (numbers[i].sec == 1 && minimNeg != MOD && change2 == -1) {
            change2 = (double)numbers[i].fir / (double)minimNeg; 
            answer2 = (answer2 * fastExpPow(minimNeg, MOD - 2LL, MOD)) % MOD;
            answer2 = (answer2 * numbers[i].fir) % MOD;
            answer2 = MOD - answer2;
        }
    }
    
    if (change1 > change2) {
        printf("%lld\n", answer1);
        return 0;
    }

    printf("%lld\n", answer2);
    return 0;
}