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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <vector>
#include <algorithm>
#define repeat(i,n) for (int i = 0; (i) < int(n); ++(i))
using ll = long long;
using namespace std;
template <class T> inline void setmin(T & a, T const & b) { a = min(a, b); }
constexpr ll inf = ll(1e18)+9;
ll solve(vector<ll> const & a, ll initial) {
    int n = a.size();
    ll cnt = abs(initial);
    ll acc = a[0] + initial;
    repeat (i,n-1) {
        ll nacc = acc + a[i+1];
        if (acc > 0 and nacc >= 0) {
            cnt += nacc+1;
            nacc = -1;
        }
        if (acc < 0 and nacc <= 0) {
            cnt += -nacc+1;
            nacc = +1;
        }
        acc = nacc;
    }
    return cnt;
}
int main() {
    int n; scanf("%d", &n);
    vector<ll> a(n); repeat (i,n) scanf("%lld", &a[i]);
    ll result = inf;
    if (a[0] != 0) setmin(result, solve(a, 0));
    if (a[0] >= 0) setmin(result, solve(a, - (a[0]+1)));
    if (a[0] <= 0) setmin(result, solve(a, + (a[0]+1)));
    printf("%lld\n", result);
    return 0;
}