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)

/// Rustで競技プログラミングの入力をスッキリ記述するマクロ
///
/// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[inline]
#[allow(dead_code)]
fn gcd(mut m: u64, mut n: u64) -> u64 {
    if m < n {
        std::mem::swap(&mut m, &mut n);
    }

    loop {
        let k = m % n;
        if k == 0 {
            return n;
        }

        m = n;
        n = k;
    }
}

#[inline]
#[allow(dead_code)]
fn lcm(m: u64, n: u64) -> u64 {
    m / gcd(m, n) * n
}

macro_rules! eprint(
    ($($arg:tt)*) => { {
        use std::io::Write;
        let r = write!(&mut ::std::io::stderr(), $($arg)*);
        r.expect("failed printing to stderr");
    } }
);

macro_rules! eprintln(
    ($($arg:tt)*) => { {
        use std::io::Write;
        let r = writeln!(&mut ::std::io::stderr(), $($arg)*);
        r.expect("failed printing to stderr");
    } }
);

macro_rules! dbg {
    ($val:expr) => {
        // Use of `match` here is intentional because it affects the lifetimes
        // of temporaries - https://stackoverflow.com/a/48732525/1063961
        match $val {
            tmp => {
                eprintln!("[{}:{}] {} = {:#?}",
                          file!(), line!(), stringify!($val), &tmp);
                tmp
            }
        }
    }
}

fn main() {
    input!{
        n: usize,
        a_v: [i64; n],
    }

    let mut maybe_answer = [0u64; 2];
    for (i, expect_positive) in [false, true].iter().enumerate() {
        let mut expect_positive = *expect_positive;
        let mut sum = 0;
        for &a in &a_v {
            sum += a;
            if expect_positive {
                if sum <= 0 {
                    maybe_answer[i] += (1 - sum) as u64;
                    sum = 1;
                }
                expect_positive = false;
            } else {
                if sum >= 0 {
                    maybe_answer[i] += (sum + 1) as u64;
                    sum = -1;
                }
                expect_positive = true;
            }
        }
    }

    let answer = maybe_answer.iter().min().unwrap();
    println!("{}", answer);
}