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)

#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define rep(i, a, b) for(i = a; i < b; i++)
#define per(i, a, b) for(i = a; i >= b; i--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define trav(x, it) for(auto it = x.begin(); it != x.end(); it++)
#define pr(x) cout << #x << " = " << x << endl
#define endl "\n"
using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector< pii > vpii;
typedef vector< pll > vpll;
const ld eps = 1e-7;
const ll mod = 1000000007;
const ll N = 998244353;

ll binpow(ll b, ll p){ ll ans = 1; while(p > 0){if(p & 1) ans = (ans * b) % mod; b = (b * b) % mod; p >>= 1;}return ans;}
ll modinv(ll x){return binpow(x, mod - 2);}

signed main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);

	ll n, k, i, ans = 1, j, lp, rp, itr;
	cin >> n >> k;
	vl a(n);
	rep(i, 0, n)
	{
		cin >> a[i];
	}
	sort(all(a));
	// rep(i, 0, n)
	// 	cout << a[i] << " ";
	// cout << endl;

    if (a[n - 1] == 0 && (k & 1))
    {
    	cout << 0 << endl;
    	return 0;
    }

    if (a[n - 1] < 0 && (k & 1))
    {
        for(i = n - 1; i >= n - k; i--)
        {
            ans = (ans * a[i]) % mod;
            while(ans < 0)
            	ans += mod;
            ans %= mod;
        }
        cout << ans << endl;
    	return 0;
    }

    i = 0;
    j = n - 1;
    if (k & 1)
    {
        ans = (ans * a[j]) % mod;
        while(ans < 0)
        	ans += mod;
        ans %= mod;
        j--; 
        k--;
    } 

    k = k / 2LL; 

    for (itr = 0; itr < k; itr++)
    {
        lp = a[i] * a[i + 1];
        rp = a[j] * a[j - 1];
        // pr(lp);
        // pr(rp);
        // cout << endl;
        if (lp > rp)
        { 
            ans = (ans * (lp % mod)) % mod; 
            while(ans < 0)
            	ans += mod;
            ans %= mod;
            i += 2; 
        } 
        else
        { 
            ans = (ans * (rp % mod)) % mod; 
            while(ans < 0)
            	ans += mod;
            ans %= mod;
            j -= 2; 
        } 
    }
    cout << ans << endl;

}