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();
        long[] A = new long[N];
        for (int i = 0; i < N; i++) {
            A[i] = sc.nextInt();
        }
        System.out.println( solve(N, A) );
    }

    private static long solve(int N, long[] A) {
        long a0 = A[0];
        if( a0 > 0 ) {
            long p = solve1(N, A, a0, 0);
            long m = solve1(N, A, -1, (int)a0 + 1);
            return Math.min(p, m);

        } else if( a0 < 0 ) {
            long p = solve1(N, A, 1, (int)a0 + 1);
            long m = solve1(N, A, a0, 0);
            return Math.min(p, m);

        } else {
            long p = solve1(N, A, 1, 1);
            long m = solve1(N, A, -1, 1);
            return Math.min(p, m);
        }
    }

    private static long solve1(int N, long[] A, long sum, long ans) {
        for (int i = 1; i < N; i++) {
            long a = A[i];

            if( sum > 0 ) {
                // 次はminusになるのを期待
                if( a + sum >= 0 ) {
                    // sumが-1になるような値にまで変更する
                    //   a + sum が 5 の場合、6 だけ操作すると -1 にできる
                    long diff = a + sum + 1;
                    ans += diff;
                    sum = -1;

                } else {
                    sum += a;
                }

            } else {
                if( a + sum <= 0 ) {
                    long diff = (a + sum) * -1 + 1;
                    ans += diff;
                    sum = 1;

                } else {
                    sum += a;
                }
            }
        }
        return ans;
    }
}