#include <iostream>
#include <stdio.h>
using namespace std;
long long a[100005];
int main(){
    int n,ans=1,m=1;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
    }
    for(int i=1;i<=n;i++){
        if(a[i]!=a[i-1]) a[m++] = a[i];
    }
    for(int i = 3;i<m;i++){
        if((a[i]-a[i-1]) * (a[i-1]-a[i-2])<0){   
            ans ++;
            i ++;
        }
        
    }
    printf("%d\n",ans);
}

### Prompt

Develop a solution in Cpp to the problem described below:

You are given an array A of length N. Your task is to divide it into several contiguous subarrays. Here, all subarrays obtained must be sorted in either non-decreasing or non-increasing order. At least how many subarrays do you need to divide A into?

Constraints

* 1 \leq N \leq 10^5
* 1 \leq A_i \leq 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 possible number of subarrays after division of A.

Examples

Input

6
1 2 3 2 2 1


Output

2


Input

9
1 2 1 2 1 2 1 2 1


Output

5


Input

7
1 2 3 2 1 999999999 1000000000


Output

3

### Response

```cpp
#include <iostream>
#include <stdio.h>
using namespace std;
long long a[100005];
int main(){
    int n,ans=1,m=1;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
    }
    for(int i=1;i<=n;i++){
        if(a[i]!=a[i-1]) a[m++] = a[i];
    }
    for(int i = 3;i<m;i++){
        if((a[i]-a[i-1]) * (a[i-1]-a[i-2])<0){   
            ans ++;
            i ++;
        }
        
    }
    printf("%d\n",ans);
}

```