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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.util.stream.Collectors;
import java.io.*;
import java.math.BigDecimal;
import java.nio.charset.StandardCharsets;

public class Main {

	public static void main(String[] args) throws IOException {
		InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);
		BufferedReader in = new BufferedReader(reader);
		Main ins = new Main(in);
		ins.calc();
		ins.showResult();
	}

	int N, K;
	long[] A;
	Deque<Long> deque;
	long MOD = 1000000007L;
	long result = 1;

	Main(BufferedReader in) throws IOException {
		String[] tokens = in.readLine().split(" ");
		N = Integer.parseInt(tokens[0]);
		K = Integer.parseInt(tokens[1]);
		A = new long[N];
		tokens = in.readLine().split(" ");
		for (int i = 0; i < N; ++i) {
			A[i] = Long.parseLong(tokens[i]);
		}
		Arrays.sort(A);
		deque = new ArrayDeque<>(N);
		for (int i = 0; i < N; ++i) {
			deque.addLast(A[i]);
		}
	}

	void calc() {
		// まず場合わけする
		if (deque.getFirst() >= 0) {
			// 大きい方からK個掛けるのがよい
			for (int i = 0; i < K; ++i) {
				result *= deque.pollLast();
				result %= MOD;
				if (result < 0) {
					result += MOD;
				}
			}
		} else if (deque.getLast() <= 0) {
			// 数が偶数か奇数かで分かれる
			if (K % 2 == 0) {
				// 絶対値が大きい方からK個掛けるのがよい
				for (int i = 0; i < K; ++i) {
					result *= deque.pollFirst();
					result %= MOD;
					if (result < 0) {
						result += MOD;
					}
				}
			} else {
				// 絶対値が小さい方からK個掛けるのがよい
				for (int i = 0; i < K; ++i) {
					result *= deque.pollLast();
					result %= MOD;
					if (result < 0) {
						result += MOD;
					}
				}
			}
		} else {
			int left = 0, right = N - 1;
			for (int i = 0; i < K / 2; i++) {
				BigDecimal leftVal = new BigDecimal(A[left]);
				leftVal = leftVal.multiply(new BigDecimal(A[left + 1]));
				BigDecimal rightVal = new BigDecimal(A[right]);
				rightVal = rightVal.multiply(new BigDecimal(A[right - 1]));
				if (rightVal.compareTo(leftVal) >= 0) {
					result *= A[right];
					result %= MOD;
					result *= A[right - 1];
					result %= MOD;
					if (result < 0) {
						result += MOD;
					}
					right -= 2;
				} else {
					result *= A[left];
					result %= MOD;
					result *= A[left + 1];
					result %= MOD;
					if (result < 0) {
						result += MOD;
					}
					left += 2;
				}
			}
			if (K % 2 == 1) {
				result *= A[right];
				result %= MOD;
				if (result < 0) {
					result += MOD;
				}
			}
		}
		System.out.println(result);
	}

	void showResult() {
	}
}