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 int long long
#define fi first
#define se second
#define otp(n) cout << n;
#define ct(n) cout << n << endl;
#define cts(n) cout << n << " ";
#define cte cout << endl;
#define ctv(v) rep(i,(v).size()) ct(v[i])
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define repo(i,o,n) for(int i=o;i<n;i++)
#define repm(i,n) for(int i=n-1;i>=0;i--)
#define repv(i,v) for(int i=0;i<(v).size();i++)
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define sperase(v,n) (v).erase(remove(all(v), n), (v).end());
#define vdelete(v) (v).erase(unique(all(v)), (v).end());
#define pb(n) push_back(n);

int n,a[100000],sum,ans,tmp;

main() {
    cin >> n;
    rep(i,n) cin >> a[i];
    rep(i,n) {
        sum+=a[i];
        if (i%2 && sum>=0) ans += sum+1,sum=-1;
        if (!(i%2) && sum<=0) ans += sum*-1+1,sum=1;
    }sum=0;
    rep(i,n) {
        sum+=a[i];
        if (i%2 && sum<=0) tmp += sum*-1+1,sum=1;
        if (!(i%2) && sum>=0) tmp += sum+1,sum=-1;
    }
    ans = min(ans,tmp);
    ct(ans)
}