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 DBG(x)                         \
    (void)(cout << "L" << __LINE__     \
                << ": " << #x << " = " \
                << (x) << '\n')
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
typedef long long ll;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const long double PI = (acos(-1));
const long double EPS = 0.0000000001;
const int INF = 0x3f3f3f3f;
const int maxn = 1100000;
const int mod = 1e9 + 7;
int a[maxn];
bool cmp(int a, int b)
{
    return abs(a) > abs(b);
}
void run_case()
{
    int n, K;
    cin >> n >> K;
    bool ok = 0;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        if (a[i] > 0)
            ok = 1;
    }
    sort(a + 1, a + 1 + n, cmp);
    ll ans = 1;
    if (ok == 0 && (K % 2 == 1))
    {
        //ll ans = 1;
        for (int i = n; i >= n - K + 1; i--)
        {
            ans = (ans * a[i] + mod) % mod;
        }
        cout << (ans + mod) % mod << "\n";
    }
    else
    {
        int cnt = 0;
        for (int i = 1; i <= K; i++)
        {
            if (a[i] < 0)
                cnt++;
            ans = (ans * a[i] + mod) % mod;
        }
        if (cnt % 2 == 0)
        {
            cout << (ans + mod) % mod << "\n";
        }
        else
        {
            //去掉一个最小的正的pos1，加一个最大的负的pos2
            //去掉一个最小的负的pos3，加一个最大的正的pos4
            int pos1 = 0, pos2 = 0, pos3 = 0, pos4 = 0;
            for (int i = K; i >= 1; i--)
            {
                if (a[i] > 0)
                {
                    pos1 = i;
                    break;
                }
            }
            for (int i = K + 1; i <= n; i++)
            {
                if (a[i] < 0)
                {
                    pos2 = i;
                    break;
                }
            }
            for (int i = K; i >= 1; i--)
            {
                if (a[i] < 0)
                {
                    pos3 = i;
                    break;
                }
            }
            for (int i = K + 1; i <= n; i++)
            {
                if (a[i] > 0)
                {
                    pos4 = i;
                    break;
                }
            }
            if (a[pos2] * a[pos3] > a[pos1] * a[pos4])
            {
                for (int i = 1; i <= n; i++)
                {
                    if (i != pos1)
                    {
                        ans = (ans * a[i] + mod) % mod;
                    }
                }
                ans = (ans * a[pos2]) % mod;
                cout << (ans + mod) % mod << "\n";
            }
            else
            {
                for (int i = 1; i <= n; i++)
                {
                    if (i != pos3)
                    {
                        ans = (ans * a[i] + mod) % mod;
                    }
                }
                ans = (ans * a[pos4]) % mod;
                cout << (ans + mod) % mod << "\n";
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setiosflags(ios::fixed) << setprecision(12);
    run_case();
    cout.flush();
    return 0;
}