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: Wrong Answer)

use std::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let a: Vec<i64> = (0..n).map(|_| input()).collect();

    let sum: Vec<i64> = a
        .iter()
        .scan(0, |acc, &a| {
            *acc += a;
            Some(*acc)
        })
        .collect();
    let mut x = 0;
    let mut ans = 0;
    for i in 1..n {
        if (sum[i - 1] + x > 0 && sum[i] + x < 0) || (sum[i - 1] + x < 0 && sum[i] + x > 0) {
            continue;
        }
        if sum[i - 1] + x > 0 {
            ans += sum[i] + x + 1;
            x -= sum[i] + x + 1;
        } else {
            ans += -sum[i] - x + 1;
            x -= sum[i] + x - 1;
        }
    }

    println!("{}", ans);
}