Disturbed People
There is a house with nn flats situated on the main street of Berlatov. Vova is watching this house every night. The house can be represented as an array of nn integer numbers a1,a2,…,ana1,a2,…,an, where ai=1ai=1 if in the ii-th flat the light is on and ai=0ai=0 otherwise.
Vova thinks that people in the ii-th flats are disturbed and cannot sleep if and only if 1<i<n1<i<n and ai−1=ai+1=1ai−1=ai+1=1 and ai=0ai=0.
Vova is concerned by the following question: what is the minimum number kk such that if people from exactly kk pairwise distinct flats will turn off the lights then nobody will be disturbed? Your task is to find this number k

Input
The first line of the input contains one integer nn (3≤n≤1003≤n≤100) — the number of flats in the house.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (ai∈{0,1}ai∈{0,1}), where aiai is the state of light in the ii-th flat.

Output
Print only one integer — the minimum number kk such that if people from exactly kk pairwise distinct flats will turn off the light then nobody will be disturbed.
Examples
Input

10
1 1 0 1 1 0 1 0 1 0
Output
2

Input
5
1 1 0 0 0
Output
0

Input

4
1 1 1 1

Output

0

Note
In the first example people from flats 22 and 77 or 44 and 77 can turn off the light and nobody will be disturbed. It can be shown that there is no better answer in this example.
There are no disturbed people in second and third examples.

正确做法：
遍历过程中，每出现一个101，可以认为前面的1与之无关，将后面的1变成0。
正确代码：

#include<cstdio>
#include<iostream>
using namespace std;
int main()
{
    int a[120],i,n,m;
    while(scanf("%d",&n)!=EOF)
    {
        m=0;
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=2;i<n;i++)
        {
            if(a[i]==0&&a[i-1]==1&&a[i+1]==1)
            {
                a[i+1]=0;
                m++;
            }
        }
        cout<<m<<'\n';
    }
    return 0;
}