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;

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[] a_sum = new int[n];

		a_sum[0] = a[0];
		for(int i=1; i<n; i++){
			a_sum[i] = a_sum[i-1] + a[i];
		}
		
		int[] a_sum_copy = new int[n];
		for(int i=0; i<n; i++){
			a_sum_copy[i] = a_sum[i];
		}

		//奇数番目までの和を正に、偶数番目までの和を負にする
		int cnt1 = 0;
		int cnt2 = 0;
		for(int i=0; i<n; i++){
			a_sum[i] += cnt1 + cnt2;
			if(i%2==0){
				if(a_sum[i]<0){
					cnt1 += (-a_sum[i]) + 1;
					a_sum[i] = 1;
				}else if(a_sum[i]==0){
					cnt1 += 1;
					a_sum[i] = 1;
				}
			}else if(i%2==1){
				if(a_sum[i]>0){
					cnt2 -= a_sum[i] + 1;
					a_sum[i] = -1;
				}else if(a_sum[i]==0){
					cnt2 -= 1;
					a_sum[i] = -1;
				}
			}
		}

		//奇数番目までの和を負に、偶数番目までの和を正にする
		int cnt3 = 0;
		int cnt4 = 0;
		for(int i=0; i<n; i++){
			a_sum_copy[i] += cnt3 + cnt4;
			if(i%2==1){
				if(a_sum_copy[i]<0){
					cnt3 += (-a_sum_copy[i]) + 1;
					a_sum_copy[i] = 1;
				}else if(a_sum_copy[i]==0){
					cnt3 += 1;
					a_sum_copy[i] = 1;
				}
			}else if(i%2==0){
				if(a_sum_copy[i]>0){
					cnt4 -= a_sum_copy[i] + 1;
					a_sum_copy[i] = -1;
				}else if(a_sum_copy[i]==0){					
					cnt4 -= 1;
					a_sum_copy[i] = -1;
				}
			}
		}

		if(cnt1-cnt2>cnt3-cnt4){
			System.out.print(cnt3-cnt4);
		}else{
			System.out.print(cnt1-cnt2);
		}

	}

}