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;
use std::io::Read;

fn main() {
    let mut stdin = io::stdin();
    let mut lines_str = String::new();
    let _ = stdin.read_to_string(&mut lines_str);
    let mut lines = lines_str.lines();
    let n :usize = lines.next().expect("").trim().parse().unwrap();
    let l2tokens = lines.next().expect("").split_whitespace();
    let mut a :Vec<isize> = Vec::new();
    for astr in l2tokens {
        a.push(astr.parse().unwrap());
    }
    let mut ans = 0;
    let mut sum = a[0];
    if sum == 0 {
        sum = 1;
        ans = 1;
    }
    for i in 1..n {
        let mut newsum = sum + a[i];
        if newsum*sum >= 0 {
            if sum > 0 {
                let diff = -1-sum-a[i];
                ans += diff.abs();
                newsum = -1;
            }else{
                let diff = 1-sum-a[i];
                ans += diff.abs();
                newsum = 1;
            }
        }
        sum = newsum;
    }
    println!("{}",ans);
}