Problem:
Score : 300 points

Problem Statement

You are given an integer sequence of length N. The i-th term in the sequence is a_i.
In one operation, you can select a term and either increment or decrement it by one.

At least how many operations are necessary to satisfy the following conditions?

For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero.

For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.

Constraints

2 ≤ n ≤ 10^5

|a_i| ≤ 10^9

Each a_i is an integer.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print the minimum necessary count of operations.

Sample Input 1

4
1 -3 1 0

Sample Output 1

4

For example, the given sequence can be transformed into 1, -2, 2, -2 by four operations. The sums of the first one, two, three and four terms are 1, -1, 1 and -1, respectively, which satisfy the conditions.

Sample Input 2

5
3 -6 4 -5 7

Sample Output 2

0

The given sequence already satisfies the conditions.

Sample Input 3

6
-1 4 3 2 -5 4

Sample Output 3

8

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;
import java.util.stream.IntStream;

public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			long[] a = new long[n];
			IntStream.range(0, n).forEach(i -> a[i] = scanner.nextLong());
			// sum1=[1,-1,...], sum2=[-1,1,...]
			int[] sum1 = new int[n], sum2 = new int[n];
			Arrays.fill(sum1, 1);
			Arrays.fill(sum2, -1);
			IntStream.range(0, n / 2).forEach(i -> {
				sum1[2 * i + 1] = -1;
				sum2[2 * i + 1] = 1;
			});
			System.out.println(Math.min(getResult(a, sum1), getResult(a, sum2)));
		}
	}

	/**
	 * @param a   数値配列
	 * @param sum 変更したい合計値の配列
	 * @return 変更すべきステップ数
	 */
	private static long getResult(final long[] a, final int[] sum) {
		int n = a.length;
		long now = 0, result = 0;
		for (int i = 0; i < n; i++) {
			now += a[i];
			if (sum[i] * now <= 0) {
				result += Math.abs(sum[i] - now);
				now = sum[i];
			}
		}
		return result;
	}
}