#include<bits/stdc++.h>
const int maxn=1e5+7;
int a[maxn];
int main()
{
	int n,cnt=1,director=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%d",&a[i]);
	for(int i=2;i<=n;i++)
	{
		if(director==0)
		{
			if(a[i-1]<a[i])
			director=1;
			if(a[i-1]>a[i])
			director=2;
		}
		else
		  if((a[i-1]<a[i]&&director==2)||(a[i-1]>a[i]&&director==1))
		    director=0,cnt++;
	}
	printf("%d",cnt);
	return 0;
}
### 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<bits/stdc++.h>
const int maxn=1e5+7;
int a[maxn];
int main()
{
	int n,cnt=1,director=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%d",&a[i]);
	for(int i=2;i<=n;i++)
	{
		if(director==0)
		{
			if(a[i-1]<a[i])
			director=1;
			if(a[i-1]>a[i])
			director=2;
		}
		else
		  if((a[i-1]<a[i]&&director==2)||(a[i-1]>a[i]&&director==1))
		    director=0,cnt++;
	}
	printf("%d",cnt);
	return 0;
}
```