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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val N = nextInt()
    val A = listOfLong()
    val B = A.toMutableList()
    var sum = A[0]
    var sign = A[0].sign()
    fun resolv(init: Long = 0L): Long {
        var ans = init
        for (n in 1 until N) {
            val now = sum + A[n]
            when (now.sign()) {
            1 ->
                if (sign != -1) {
                    ans += (1 + now)
                    sum = -1
                } else sum = now
            -1 ->
                if (sign != 1) {
                    ans += (1 - now)
                    sum = 1
                } else sum = now
            0 ->
                if (sign == 1) {
                    ans += 1
                    sum = -1
                } else {
                    ans += 1
                    sum = 1
                }
            }
            sign = sum.sign()
        }
        return ans
    }
    val ans1 = resolv()
    sum = A[0]
    sign = A[0].sign()
    val ans2 =
        if (sign == -1) {
            sum = 1
            sign = 1
            resolv((1 - A[0]))
        } else {
            sum = -1
            sign = -1
            resolv((1 + A[0]))
        }
    println(Math.min(ans1, ans2))
}
fun Long.sign() = if (this > 0) 1 else if (this < 0) -1 else 0

fun next() = readLine()!!
fun nextInt(delta: Int = 0) = Integer.parseInt(next()) + delta
fun listOfString() = next().split(" ")
fun listOfInt() = listOfString().map(String::toInt)
fun listOfLong() = listOfString().map(String::toLong)