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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

#define SIZE_NUM 100000
#define FLG_INIT 0
#define FLG_PLUS 1
#define FLG_MINUS -1


int main(void)
{
	/* 変数宣言 */
	int iNum[SIZE_NUM] = {0};	// 入力データ
	int iNumber = 0;			// 入力データ個数
	int iLoop;					// ループカウンタ
	int iSum = 0;				// 和
	int iFlg = FLG_INIT;		// プラスマイナスフラグ
	int iResult = 0;			// 操作回数
	
	/* データ読み込み */
	scanf("%d", &iNumber);
	
	for(iLoop = 0; iLoop < iNumber; iLoop++)
	{
		scanf("%d", &iNum[iLoop]);
	}
	
	/* 操作回数計算 */
	
	for(iLoop = 0; iLoop < iNumber; iLoop++)
	{
		/* 最初から0が続く場合だけ特別処理 */
		if(iFlg == FLG_INIT && iNum[iLoop] == 0)
		{
			if(iLoop == 0)
			{
				iResult++;
			}
			else
			{
				iResult += 2;
			}
			continue;
		}
		
		/* 和を算出 */
		iSum += iNum[iLoop];
		
		/* フラグが立ってない場合、フラグを立てて次のループへ */
		if(iFlg == FLG_INIT)
		{
			if(iNum[iLoop] > 0)
			{
				iFlg = FLG_PLUS;
			}
			else
			{
				iFlg = FLG_MINUS;
			}
			continue;
		}
		
		/* フラグ(1 or -1)と当該ループまでの和の符号確認 */
		/* 積の符号が同じまたは0だとだめ */
		if(iFlg * iSum >= 0)
		{
			iResult += abs(iSum) + 1;
			iSum = iFlg * -1;	// フラグと異なる符号の絶対値1の数
		}
		iFlg *= -1;			// フラグの符号を反転させる
	}
	
	/* 結果出力 */
	printf("%d\n", iResult);
	
	return 0;
}