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.io.IOException;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws IOException{
		Sequence solver = new Sequence();
		solver.readInput();
		solver.solve();
		solver.writeOutput();
	}


	static class Sequence {
		private int n;
		private long a[];

		private long output;
		private Scanner scanner;

		public Sequence() {
			this.scanner = new Scanner(System.in);
		}

		public void readInput() {
			n = Integer.parseInt(scanner.next());
			a = new long[n];
			for(int i=0; i<n; i++) {
				a[i] = Integer.parseInt(scanner.next());
			}
		}

		private int count(int sign) {
			int count=0;
			long sum=0;
			for(int i=0; i<n; i++) {
				sum += a[i];
				if(i%2==sign) {
					// a[i]までの合計を正にするとき
					if(sum<=0) {
						count += 1-sum;
						sum = 1;
					}
				} else {
					// a[i]までの合計を負にするとき
					if(0<=sum) {
						count += 1+sum;
						sum = -1;
					}
				}
			}
			return count;
		}

		public void solve() {

			/* https://atcoder.jp/contests/abc059/submissions/8183898 */
			long sum1 = 0;
			long count1 = 0;
			long sum2 = 0;
			long count2 = 0;
			for (int i = 0; i < n; i++) {
			    sum1 += a[i];
			    sum2 += a[i];
			    if (i % 2 == 0) {
			        if (sum1 <= 0) {
			            count1 += 1 - sum1;
			            sum1 = 1;
			        }
			        if (sum2 >= 0) {
			            count2 += sum2 + 1;
			            sum2 = -1;
			        }
			    } else {
			        if (sum2 <= 0) {
			            count2 += 1 - sum2;
			            sum2 = 1;
			        }
			        if (sum1 >= 0) {
			            count1 += sum1 + 1;
			            sum1 = -1;
			        }
			    }
			}
			output = Math.min(count1,count2);
		}

		public void writeOutput() {
			System.out.println(output);
		}
	}

}