#include <cstdio>
#define REP(i,n) for (int i=0;i<(n);i++)
int main(){
int N;
int count=1;
int v=0;
int A[114514]={};
scanf("%d\n",&N);
REP(i,N){scanf("%d",&A[i]);}
if(N==1){printf("1");return 0;}

REP(i,N-1){
if(A[i]<A[i+1]){
if(v==-1){
//printf("%d\n",i);
count++;
//i++;
v=0;
}else{
v=1;
}
}

else if(A[i]>A[i+1]){
if(v==1){
//printf("%d\n",i);
count++;
//i++;
v=0;
}else{
v=-1;
}
}


}
printf("%d\n",count);
return 0;
}
### 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 <cstdio>
#define REP(i,n) for (int i=0;i<(n);i++)
int main(){
int N;
int count=1;
int v=0;
int A[114514]={};
scanf("%d\n",&N);
REP(i,N){scanf("%d",&A[i]);}
if(N==1){printf("1");return 0;}

REP(i,N-1){
if(A[i]<A[i+1]){
if(v==-1){
//printf("%d\n",i);
count++;
//i++;
v=0;
}else{
v=1;
}
}

else if(A[i]>A[i+1]){
if(v==1){
//printf("%d\n",i);
count++;
//i++;
v=0;
}else{
v=-1;
}
}


}
printf("%d\n",count);
return 0;
}
```