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>
#include <type_traits>
using namespace std;

using ll = int64_t;
#define int ll

#define FOR(i, a, b) for (int i = int(a); i < int(b); i++)
#define REP(i, b) FOR(i, 0, b)
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define ALL(x) x.begin(), x.end()

using pi = pair<int, int>;
using vi = vector<int>;
using ld = long double;

template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}

template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  REP(i, (int)v.size()) {
    if (i)
      os << ",";
    os << v[i];
  }
  os << "}";
  return os;
}

ll read() {
  ll i;
  scanf("%" SCNd64, &i);
  return i;
}

void printSpace() { printf(" "); }

void printEoln() { printf("\n"); }

void print(ll x, int suc = 1) {
  printf("%" PRId64, x);
  if (suc == 1)
    printEoln();
  if (suc == 2)
    printSpace();
}

string readString() {
  static char buf[3341000];
  scanf("%s", buf);
  return string(buf);
}

char *readCharArray() {
  static char buf[3341000];
  static int bufUsed = 0;
  char *ret = buf + bufUsed;
  scanf("%s", ret);
  bufUsed += strlen(ret) + 1;
  return ret;
}

template <class T, class U> void chmax(T &a, U b) {
  if (a < b)
    a = b;
}

template <class T, class U> void chmin(T &a, U b) {
  if (b < a)
    a = b;
}

template <class T> T Sq(const T &t) { return t * t; }

const int mod = 1e9 + 7;

signed main() {
  int n = read(), sum = 0, ans = 0;
  int a = read();
  bool isPlus = (a > 0);
  sum = a;
  REP(i, n - 1) {
    sum += read();
    if (isPlus && sum >= 0) {
      ans += sum + 1;
      sum = -1;
    } else if (!isPlus && sum <= 0) {
      ans += -1 * sum + 1;
      sum = 1;
    }
    isPlus ^= 1;
  }
  print(ans);
}