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 (1.15.1) (judge verdict: Accepted)

use std::cmp::*;
use std::io::*;

fn main() {
    let mut s: String = String::new();
    stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let mut a: Vec<i64> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();
    for i in 0..n - 1 {
        a[i + 1] += a[i];
    }

    let mut ans1 = 0;
    let mut ans2 = 0;
    let mut diff = 0i64;
    for i in 0..n {
        if i % 2 == 0 && a[i] + diff <= 0 {
            // sei
            ans1 += 1 - (a[i] + diff);
            diff += 1 - (a[i] + diff);
        }
        if i % 2 == 1 && a[i] + diff >= 0 {
            // hu
            ans1 += a[i] + diff + 1;
            diff -= a[i] + diff + 1;
        }
    }
    diff = 0;
    for i in 0..n {
        if i % 2 == 0 && a[i] + diff >= 0 {
            // sei
            ans2 += a[i] + diff + 1;
            diff -= a[i] + diff + 1;
        }
        if i % 2 == 1 && a[i] + diff <= 0 {
            // hu
            ans2 += 1 - (a[i] + diff);
            diff += 1 - (a[i] + diff);
        }
    }
    println!("{}", min(ans1, ans2));
}