#include <bits/stdc++.h>
using namespace std;
int main(){
	int n,s=0;
	scanf("%d",&n);
	long long int a[n];
	for(int i=0;i<n;++i){
		int t;
		scanf("%d",&t);
		if(i!=0&&t==a[s-1]){
			continue;
		}
		else {
			a[s]=(long long )t;
			++s;
		}
	}
	int res=1;
	for(int i=1;i<s-1;){

		if(a[i]-a[i-1] < 0 && a[i+1]-a[i]>0 || a[i]-a[i-1] > 0 && a[i+1]-a[i] < 0){
			++res;
			i+=2;
 
		}
		else ++i;
 
	}
	printf("%d",res);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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 <bits/stdc++.h>
using namespace std;
int main(){
	int n,s=0;
	scanf("%d",&n);
	long long int a[n];
	for(int i=0;i<n;++i){
		int t;
		scanf("%d",&t);
		if(i!=0&&t==a[s-1]){
			continue;
		}
		else {
			a[s]=(long long )t;
			++s;
		}
	}
	int res=1;
	for(int i=1;i<s-1;){

		if(a[i]-a[i-1] < 0 && a[i+1]-a[i]>0 || a[i]-a[i-1] > 0 && a[i+1]-a[i] < 0){
			++res;
			i+=2;
 
		}
		else ++i;
 
	}
	printf("%d",res);
	return 0;
}
```