time limit per test

1 second

memory limit per test

256 megabytes

Chaneka, Pak Chanek's child, is an ambitious kid, so Pak Chanek gives her the following problem to test her ambition.

Given an array of integers?[A1,A2,A3,…,AN][A1,A2,A3,…,AN]. In one operation, Chaneka can choose one element, then increase or decrease the element's value by?11. Chaneka can do that operation multiple times, even for different elements.

What is the minimum number of operations that must be done to make it such that?A1×A2×A3×…×AN=0A1×A2×A3×…×AN=0?

Input

The first line contains a single integer?NN?(1≤N≤1051≤N≤105).

The second line contains?NN?integers?A1,A2,A3,…,ANA1,A2,A3,…,AN?(?105≤Ai≤105?105≤Ai≤105).

Output

An integer representing the minimum number of operations that must be done to make it such that?A1×A2×A3×…×AN=0A1×A2×A3×…×AN=0.

Examples

Input

Copy

3
2 -6 5

Output

Copy

2

Input

Copy

1
-3

Output

Copy

3

Input

Copy

5
0 -1 0 1 0

Output

Copy

0

Note

In the first example, initially,?A1×A2×A3=2×(?6)×5=?60A1×A2×A3=2×(?6)×5=?60. Chaneka can do the following sequence of operations:

1. Decrease the value of?A1A1?by?11. Then,?A1×A2×A3=1×(?6)×5=?30A1×A2×A3=1×(?6)×5=?30
2. Decrease the value of?A1A1?by?11. Then,?A1×A2×A3=0×(?6)×5=0A1×A2×A3=0×(?6)×5=0

In the third example, Chaneka does not have to do any operations, because from the start, it already holds that?A1×A2×A3×A4×A5=0×(?1)×0×1×0=0

解題說明：水題，找到絕對值最小的數，然后將該數變成0即可。

#include<stdio.h>
int main()
{int n, x, ans = 1e5;scanf("%d", &n);for (int i = 0; i < n; ++i){scanf("%d", &x);if (abs(x) <= ans){ans = abs(x);}}printf("%d", ans);return 0;
}