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: Time Limit Exceeded)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n];
		long[] s1 = new long[n];

		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}

		long sum1 = 0;
		s1[0] = a[0];
		// - + - + となる場合
		while (s1[0] >= 0) {
			s1[0]--;
			sum1++;
		}
		for (int i = 1; i < n; i++) {
			s1[i] = s1[i-1] + a[i];
			// iが奇数の場合は正にする
			if (i % 2 != 0) {
				if (s1[i-1]*s1[i] >= 0) {
					sum1 = sum1 + 1 - s1[i];
					s1[i] = 1;
				}
			}
			// iが偶数の場合は負にする
			else {
				if (s1[i-1]*s1[i] >= 0) {
					sum1 = sum1 + 1 + s1[i];
					s1[i] = -1;
				}
			}
		}

		Arrays.fill(s1, 0);
		long sum2 = 0;
		s1[0] = a[0];
		// + - + - となる場合
		while (s1[0] <= 0) {
			s1[0]++;
			sum2++;
		}
		for (int i = 1; i < n; i++) {
			s1[i] = s1[i-1] + a[i];
			// iが奇数の場合は負にする
			if (i % 2 != 0) {
				if (s1[i-1]*s1[i] >= 0) {
					sum2 = sum2 + 1 + s1[i];
					s1[i] = -1;
				}
			}
			// iが偶数の場合は正にする
			else {
				if (s1[i-1]*s1[i] >= 0) {
					sum2 = sum2 + 1 - s1[i];
					s1[i] = 1;
				}
			}
		}

		System.out.println(Math.min(sum1, sum2));
	}

}