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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
 
int main(){
	unsigned long long N;
	long long *a;
	long long *sum, *sum_;
	unsigned long long ans1 = 0, ans2 = 0;
	scanf("%d", &N);
	a = (int *)malloc(sizeof(long long) * N);
	sum = (int *)malloc(sizeof(long long) * N);
	sum_ = (int *)malloc(sizeof(long long) * N);
	for(unsigned long long i = 0; i < N; i++) scanf("%d", a + i);
	for(unsigned long long i = 0; i < N; i++) sum[i] = 0;
	sum[0] = a[0];
	for(unsigned long long i = 1; i < N; i++) sum[i] = sum[i - 1] + a[i];
	for(unsigned long long i = 0; i < N; i++) sum_[i] = sum[i];
 
	for(unsigned long long i = 0; i < N; i++){
		if(i%2){if(sum[i] >= 0){
			ans1 += sum[i]; ans1++;
			for(unsigned long long j = i + 1; j < N; j++){
				sum[j] -= sum[i]; sum[j]--;
			}
		}}else if(sum[i] <= 0){
			ans1 -= sum[i]; ans1++;
			for(unsigned long long j = i + 1; j < N; j++){
				sum[j] -= sum[i]; sum[j]++;
			}
		}
	}
	for(unsigned long long i = 0; i < N; i++){
		if(i%2){if(sum_[i] <= 0){
			ans2 -= sum_[i]; ans2++;
			for(unsigned long long j = i + 1; j < N; j++){
				sum_[j] -= sum_[i]; sum_[j]++;
			}
		}}else if(sum_[i] >= 0){
			ans2 += sum_[i]; ans2++;
			for(unsigned long long j = i + 1; j < N; j++){
				sum_[j] -= sum_[i]; sum_[j]--;
			}
		}
	}
	printf("%d\n", (ans1 < ans2) ? ans1 : ans2);
}