#include <bits/stdc++.h>
using namespace std;
#define ll long long

int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<ll> a(N);
	for(int i=0;i<N;i++)cin >> a[i];
	ll ans = 0;
	for(int i=0;i<N;i++){
		while(i+1<N && a[i]==a[i+1])i++;
		if(i+1<N&&a[i]<a[i+1]){
			while(i+1<N&&a[i]<=a[i+1])i++;
		}else if(i+1<N&&a[i]>a[i+1]){
			while(i+1<N&&a[i]>=a[i+1])i++;
		}
		ans++;
	}
	cout << ans << endl;
}
### 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;
#define ll long long

int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<ll> a(N);
	for(int i=0;i<N;i++)cin >> a[i];
	ll ans = 0;
	for(int i=0;i<N;i++){
		while(i+1<N && a[i]==a[i+1])i++;
		if(i+1<N&&a[i]<a[i+1]){
			while(i+1<N&&a[i]<=a[i+1])i++;
		}else if(i+1<N&&a[i]>a[i+1]){
			while(i+1<N&&a[i]>=a[i+1])i++;
		}
		ans++;
	}
	cout << ans << endl;
}
```