Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

6
1 2 3 2 2 1

Sample Output 1

2

One optimal solution is to divide the array into [1,2,3] and [2,2,1].

Sample Input 2

9
1 2 1 2 1 2 1 2 1

Sample Output 2

5

Sample Input 3

7
1 2 3 2 1 999999999 1000000000

Sample Output 3

3

Solution in C (Clang 3.8.0) (judge verdict: Wrong Answer)

// 161-C Replacing Integer

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <math.h>
#include <assert.h>
#include <string.h>
#include <memory.h>
#include <malloc.h>
#include <stdlib.h>

// qsortで用いる比較用関数．*a<*bならaが先になる．
int compare(const void* a, const void* b) {
    return -1 * (*(int*)a - *(int*)b);
}


int main() {
    int n,i,cnt=1,diff;
    int* a;
    int isReset = 0;

    scanf("%d", &n);
    if (n == 1) {
        printf("1");
        return 0;
    }

    a = (int*)malloc(sizeof(int) * n);
    for (i = 0; i < n - 1; i++) {
        scanf("%d", &a[i]);
    }

    diff = a[i] - a[i + 1];

    for (i = 1; i < n - 1; i++) {

        if (diff == 0 && a[i] == a[i + 1]) continue;

        if (isReset == 0) {
            if (diff < 0 && a[i] > a[i + 1]) { // 一つ前のiについて，a[i] < a[i+1]
                cnt++;
                isReset = 1;
            }

            if (diff > 0 && a[i] < a[i + 1]) {
                cnt++;
                isReset = 1;
            }
            
        }
        else {
            isReset = 0;
        }
        diff = a[i] - a[i + 1];
    }


    printf("%d", cnt);
    return 0;
}