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 Rust (1.42.0) (judge verdict: Accepted)

// -*- coding:utf-8-unix -*-

use proconio::{input, fastout};

// ABC086C - Traveling
// https://atcoder.jp/contests/abs/fasks/arc089_a

#[fastout]
fn main() {
    input! {
        n: usize,
        a_vec: [isize; n],
    }

    // Pattern 1: resulting a_1 is positive
    let mut count1 = 0;
    let mut prevsum;
    if a_vec[0] <= 0 {
        count1 += a_vec[0].abs() + 1;
        prevsum = 1;
    } else {
        prevsum = a_vec[0];
    }
    let mut nowsum;
    for i in 1..n {
        nowsum = prevsum + a_vec[i];
        if nowsum*prevsum < 0 {
            prevsum = nowsum;
            continue;
        }
        if prevsum > 0 {
            // Make nowsum negative
            count1 += nowsum.abs() + 1;
            prevsum = -1;
        } else {// prevsum < 0
            // Make nowsum positive
            count1 += nowsum.abs() + 1;
            prevsum = 1;
        }
    }

    // Pattern 2: resulting a_1 is negative
    let mut count2 = 0;
    let mut prevsum;
    if a_vec[0] >= 0 {
        count2 += a_vec[0].abs() + 1;
        prevsum = -1;
    } else {
        prevsum = a_vec[0];
    }
    let mut nowsum;
    for i in 1..n {
        nowsum = prevsum + a_vec[i];
        if nowsum*prevsum < 0 {
            prevsum = nowsum;
            continue;
        }
        if prevsum > 0 {
            // Make nowsum negative
            count2 += nowsum.abs() + 1;
            prevsum = -1;
        } else {// prevsum < 0
            // Make nowsum positive
            count2 += nowsum.abs() + 1;
            prevsum = 1;
        }
    }

    println!("{}", count1.min(count2));
}