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: Wrong Answer)

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];
		for(int i = 0 ; i < n ; i++) a[i] = sc.nextInt();
		int sum = 0, ans = 0, ans2 = 0;
		// 偶数位置までの和：正、奇数位置までの和：負
		for(int i = 0 ; i < n ; i++) {
			if(i % 2 == 0) {
				if(sum + a[i] >= 0) {
					ans += sum + a[i] + 1;
					sum = -1;
				} else {
					sum += a[i];
				}
			} else {
				if(sum + a[i] <= 0) {
					ans += 1 - (sum + a[i]);
					sum = 1;
				} else {
					sum += a[i];
				}
			}
		}
		sum = 0;
		// 偶数位置までの和：正、奇数位置までの和：負
		for(int i = 0 ; i < n ; i++) {
			if(i % 2 == 1) {
				if(sum + a[i] >= 0) {
					ans2 += sum + a[i] + 1;
					sum = -1;
				} else {
					sum += a[i];
				}
			} else {
				if(sum + a[i] <= 0) {
					ans2 += 1 - (sum + a[i]);
					sum = 1;
				} else {
					sum += a[i];
				}
			}
		}
		System.out.println(ans2);
		System.out.println(Math.min(ans, ans2));
	}
}