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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES
#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <utility>
#include <algorithm>
#include <queue>
#include <list>
#include <stack>
#include <math.h>
#include <map>
#include <unordered_map>
#include <set>
using namespace std;

#define ll long long
#define ull unsigned long long

const ll limit = (ll)(1e9 + 7);

ll gcd(ll a, ll b) {
	if (a < b)
		swap(a, b);
	while (ll r = a % b)
		a = b, b = r;
	return b;
}

class mod
{
public:
	ll val;

	mod(ll v)
	{
		val = v % limit;
	}

	mod()
	{
		val = 0;
	}

	const mod operator -(const mod &v) const
	{
		mod ret(v.val);
		ret -= val;
		return ret;
	}

	const mod operator +(const mod &v) const
	{
		mod ret(v.val);
		ret += val;
		return ret;
	}

	void operator -=(const mod &v)
	{
		val -= v.val;

		if (val < 0)
			val += limit;
	}

	void operator +=(const mod &v)
	{
		if (val + v.val >= limit)
			val -= limit;
		val += v.val;
	}

	void operator +=(ll &v)
	{
		if (val + v >= limit)
			val -= limit;
		val += v;
	}

	void operator *=(const mod &v)
	{
		val *= v.val;
		val %= limit;
	}

	void operator *=(ll &v)
	{
		val *= v;
		val %= limit;
	}

	ll modinv(ll a)
	{
		ll b = limit, u = 1, v = 0;
		while (b)
		{
			ll t = a / b;
			a -= t * b; swap(a, b);
			u -= t * v; swap(u, v);
		}
		u %= limit;
		if (u < 0)
			u += limit;
		return u;
	}

	void operator /=(const mod &v)
	{
		val *= modinv(v.val);
		val %= limit;
	}

	void operator /=(ll &v)
	{
		val *= modinv(v);
		val %= limit;
	}
};

struct UnionFind
{
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
	vector<int> childN;
	ll r = 0;

	UnionFind(int N) : par(N), childN(N)
	{
		//最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++)
			par[i] = i, childN[i] = 1;
	}

	int root(int x)
	{
		// データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y)
	{
		// xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
		r -= childN[rx] * (childN[rx] - 1) / 2;
		r -= childN[ry] * (childN[ry] - 1) / 2;
		childN[ry] += childN[rx];
		childN[rx] = 0;
		r += childN[ry] * (childN[ry] - 1) / 2;
	}

	bool same(int x, int y)
	{
		// 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};

class WarshallFloyd
{
public:
	vector<vector<int>> d;
	int n;
	
	WarshallFloyd(int _n) : n(_n) { d.resize(n, vector<int>(n, limit)); }

	void run()
	{
		for (int k = 0; k < n; k++)
			for (int i = 0; i < n; i++)
				for (int j = 0; j < n; j++)
					if (d[i][k] < limit && d[k][i] < limit)
						d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
	}

};

//	for priority queue
//	default -> 	priority_queue<int, std::vector<int>, std::greater<int>> q;
//	priority_queue<ipair, vector<ipair>, greater<ipair>>
using ipair = pair<int, pair<int, int>>;
bool operator>(const ipair a, const ipair b) { return a.first > b.first; };
bool operator<(const ipair a, const ipair b) { return a.first < b.first; };

using xpair = pair<int, int>;
bool operator>(const xpair &a, const xpair &b) { return a.second > b.second || (a.second == b.second && a.first > b.first); };
bool operator<(const xpair &a, const xpair &b) { return a.second < b.second || (a.second == b.second && a.first > b.first); };
bool operator==(const xpair &a, const xpair &b) { return a.first == b.first; };

int main()
{
	int n, k;
	cin >> n >> k;
	vector<ll> a(n, 0), b(n, 0);
	int na = 0, nb = 0;
	mod r(1);
	for (int i = 0; i < n; i++)
	{
		ll in;
		cin >> in;
		if (in >= 0)
			a[na++] = in;
		else
			b[nb++] = in;
	}

	sort(a.begin(), a.end(), greater<int>());

	sort(b.begin(), b.end());

	if (na == 0)
	{
		if (k % 2 == 1)
		{
			for (int i = nb - 1, j = 0; j < k; i--, j++)
				r *= -b[i];
			printf("%lld", limit - r.val);
			return 0;
		}

		for (int i = 0; i < k; i++)
			r *= -b[i];
		printf("%lld", r.val);
		return 0;
	}

	//	1以上の数がないとき
	if (a[0] == 0)
	{
		if (k % 2 == 1 || nb < k)
		{
			printf("0");
			return 0;
		}
		for (int i = 0; i < k; i++)
			r *= -b[i];
		printf("%lld", r.val);
		return 0;
	}

	//	使うaの数
	int ia = max(0, k - nb);
	if ((k - ia) % 2 == 1)
		ia++;
	int maxi = ia;
	ia += 2;
	double rate = 1.0f;
	for (int i = ia; i <= na && i <= k; i += 2)
	{
		if (a[i - 1] == 0 || a[i - 2])
			break;
		rate *= a[i - 1];
		rate /= b[k - i + 1];
		rate *= a[i - 2];
		rate /= b[k - i];
		if (rate <= 1.0f)
			continue;
		rate = 1.0f;
		maxi = i;
	}

	for (int i = 0; i < maxi; i++)
		r *= a[i];
	for (int i = 0; i < k - maxi; i++)
		r *= -b[i];
	printf("%lld", r.val);
	return 0;
}