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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, e) for(int (i) = 0; (i) < (e); ++(i))
#define FOR(i, b, e) for(int (i) = (b); (i) < (e); ++(i))
#define ALL(c) (c).begin(), (c).end()
#define PRINT(x) cout << (x) << "\n"
using namespace std;
using ll = long long; using pint = pair<int, int>; using pll = pair<ll, ll>;
const long long MOD = 1000000007;

ll N, a[100010];

ll solve(ll z){
    ll ans = !(a[0] == z), sum = a[0];
    a[0] = z;
    FOR(i, 1, N){
        if(sum > 0){
            ans += max(0LL, sum + a[i] + 1);
            sum += a[i] - max(0LL, sum + a[i] + 1);
        }else if(sum < 0){
            ans += max(0LL, -a[i] - sum + 1);
            sum += a[i] + max(0LL, -a[i] - sum + 1);
        }
    }
    return ans;
}

signed main(){
    cin >> N;
    REP(i, N) cin >> a[i];
    if(a[0] == 0) PRINT(min(solve(1), solve(-1)));
    else PRINT(solve(a[0]));
    return 0;
}