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 rep(i,n) for(int i = 0; i < (n); ++i)
#define srep(i,s,t) for(int i = s; i < t; ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
#define yn {puts("Yes");}else{puts("No");}
#define MAX_N 200005

int main() {
    int n, k;
    cin >> n >> k;
    ll a[n];
    rep(i,n)cin >> a[i];
    vector<ll> plus, minus;
    rep(i,n){
        if(a[i] > 0)plus.push_back(a[i]);
        else minus.push_back(a[i]);
    }
    n = plus.size();
    int m = minus.size();
    sort(plus.begin(), plus.end(), greater<ll>());
    sort(minus.begin(), minus.end());

    ll ans = 1;
    ll MOD = 1e9+7;

    if(n == 0){
        if(k % 2 == 0){
            rep(i,k){
                ans *= abs(minus[i]);
                ans %= MOD;
            }
        }else{
            rep(i,k){
                ans *= abs(minus[m-1-i]);
                ans %= MOD;
            }
            ans = MOD - ans;
        }
    }

    else if(k == n+m){
        rep(i,n){
            ans *= plus[i];
            ans %= MOD;
        }
        rep(i,m){
            ans *= MOD + minus[i];
            ans %= MOD;
        }
    }

    else{ // 最初にn,m,kをすべて偶数にしてしまう.
        if(k % 2 == 0){
            if(n % 2 == 1){
                plus.erase(plus.begin() + n - 1);
                n--;
            }
            if(m % 2 == 1){
                minus.erase(minus.begin() + m - 1);
                m--;
            }
        }else{
            ans *= plus[0];
            plus.erase(plus.begin());
            n--;
            k--;
            if(n % 2 == 1){
                plus.erase(plus.begin() + n - 1);
                n--;
            }
            if(m % 2 == 1){
                minus.erase(minus.begin() + m - 1);
                m--;
            }
        }
        vector<ll> v;
        rep(i,n/2){
            v.push_back(plus[i*2] * plus[i*2+1]);
        }
        rep(i,m/2){
            v.push_back(minus[i*2] * minus[i*2+1]);
        }
        sort(v.begin(), v.end(), greater<ll>());
        rep(i,k/2){
            ans *= v[i] % MOD;
            ans %= MOD;
        }
    }


    cout << ans << endl;
    return 0;
}