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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
using namespace std;

#define ll         long long
#define PI         acos(-1.0)
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)

//方針
//i, i+1は前から調整していくしかない富豪がことなったら0
//最後で和の調整

ll llabs(ll x){
    return x>=0?x:-x;
}

bool hugou(ll x){
    return x>=0?true:false;
}

int main(){
    int n;
    cin >> n;
    ll sum = 0;
    ll ans = 0;
    ll A[n];
    FOR(i, 0, n) cin >> A[i];
    //000000注意
    int j = 0;
    while(A[j]==0){
        j++;
    }
    ans += (j-1) * 2 + 1;
    if(A[j]>0) sum = -1;
    else sum = 1;
    FOR(i, j, n){
        ll a = A[i];
        if(sum+a==0){
            ans++;
            if(sum<0) sum = 1;
            else sum = -1;
        }
        else if(!(hugou(sum)^hugou(sum+a))){
            ans += llabs(sum+a) + 1;
            if(sum<0) sum = 1;
            else sum = -1;
        }
        else sum += a;
    }
    if(sum==0) ans++;
    cout << ans << endl;
}