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: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
typedef long long LL;
typedef pair<int, int> PII;

int n;
int a[100000];

int main() {
    scanf("%d", &n);
    forn(i, n) scanf("%d", a + i);
    LL ans = 0, sum = 0;
    forn(i, n) {
        sum += a[i];
        if (i & 1) {
            if (sum > -1) {
                ans += sum + 1;
                sum = -1;
            }
        } else {
            if (sum < 1) {
                ans += 1 - sum;
                sum = 1;
            }
        }
    }
    LL ans2 = ans;
    ans = sum = 0;
    forn(i, n) {
        sum += a[i];
        if (!(i & 1)) {
            if (sum > -1) {
                ans += sum + 1;
                sum = -1;
            }
        } else {
            if (sum < 1) {
                ans += 1 - sum;
                sum = 1;
            }
        }
    }
    cout << min(ans, ans2) << endl;
    return 0;
}