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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (ll (i)=(0);(i)<(ll)(n);++(i))
typedef pair<int, int> P;
typedef long long ll;

int main() {
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    rep(i, n) cin >> a[i], b[i] = a[i];

    ll sum = a[0];
    ll ans = 0;
    if (sum == 0) sum++, ans++;

    for (int i=1; i<n; ++i) {
        if ( (sum > 0 and sum+a[i] < 0) or (sum < 0 and sum+a[i] > 0) ) {
            sum += a[i];
        }
        else {
            if (sum > 0) {
                sum += a[i];
                for (; sum>=0; --sum) {
                    ++ans;
                }
            }
            else {
                sum += a[i];
                for (; sum<=0; ++sum) {
                    ++ans;
                }
            }
        }
    }

    ll ans2 = 0;
    sum = a[0];
    if (sum > 0) {
        for (; sum>=0; --sum) {
            ++ans2;
        }
    }
    else {
        for (; sum<=0; ++sum) {
            ++ans2;
        }
    }
    if (sum == 0) sum--, ans2++;
    
    for (int i=1; i<n; ++i) {
        if ( (sum > 0 and sum+a[i] < 0) or (sum < 0 and sum+a[i] > 0) ) {
            sum += a[i];
        }
        else {
            if (sum > 0) {
                sum += a[i];
                for (; sum>=0; --sum) {
                    ++ans2;
                }
            }
            else {
                sum += a[i];
                for (; sum<=0; ++sum) {
                    ++ans2;
                }
            }
        }
    }

    cout << min(ans, ans2) << endl;
}