#include<stdio.h> 
int main()
{
	long long z[100005],i,j,k=0,l=0,t=0,n;
	scanf("%lld",&n);
	for(i=0;i<n;i++)
	scanf("%lld",&z[i]);
	for(i=0;i<n-1;i++)
	{
		if(k==0)
		{
			for(j=i+1;j<n;j++)
			{
				if(z[i]!=z[j])
				{
					if(z[j]>z[i])
						k=1;
					else
						k=-1;
					i=j-1;
					break;
				}
			}
		}
		else
		{
			if(k==1&&z[i+1]<z[i])
				{l++;k=0;t=1;}
			if(k==-1&&z[i+1]>z[i])
			{	l++;k=0;t=1;}
		}
	}
	
	printf("%lld\n",l+1);
}
### Prompt

Generate a cpp solution to the following problem:

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<stdio.h> 
int main()
{
	long long z[100005],i,j,k=0,l=0,t=0,n;
	scanf("%lld",&n);
	for(i=0;i<n;i++)
	scanf("%lld",&z[i]);
	for(i=0;i<n-1;i++)
	{
		if(k==0)
		{
			for(j=i+1;j<n;j++)
			{
				if(z[i]!=z[j])
				{
					if(z[j]>z[i])
						k=1;
					else
						k=-1;
					i=j-1;
					break;
				}
			}
		}
		else
		{
			if(k==1&&z[i+1]<z[i])
				{l++;k=0;t=1;}
			if(k==-1&&z[i+1]>z[i])
			{	l++;k=0;t=1;}
		}
	}
	
	printf("%lld\n",l+1);
}
```