#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int a[maxn],n;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	int ans=0;
	for(int i=1,j;i<=n;i=j+1){
		j=i;
		ans++;
		while(j+1<=n&&a[i]==a[j+1])j++;
		if(j+1<=n){
			if(a[i]<a[j+1]){
				while(j+1<=n&&a[j]<=a[j+1])j++;
			}else{	
				while(j+1<=n&&a[j]>=a[j+1])j++;
			}	
		}
	}
	cout<<ans<<endl;
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int a[maxn],n;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	int ans=0;
	for(int i=1,j;i<=n;i=j+1){
		j=i;
		ans++;
		while(j+1<=n&&a[i]==a[j+1])j++;
		if(j+1<=n){
			if(a[i]<a[j+1]){
				while(j+1<=n&&a[j]<=a[j+1])j++;
			}else{	
				while(j+1<=n&&a[j]>=a[j+1])j++;
			}	
		}
	}
	cout<<ans<<endl;
	return 0;
}
```