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# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
public struct Ant
{
    public int pos;
    public bool isClockWise;
}
class Ants
{
    static void Main(string[] args)
    {
        string[] input = Console.ReadLine().Split(' ');
        int n = int.Parse(input[0]);
        int l = int.Parse(input[1]);
        int t = int.Parse(input[2]);
        Ant[] a = new Ant[n];
        for (int i = 0; i < n; i++)
        {
            string[] s = Console.ReadLine().Split(' ');
            a[i].pos = int.Parse(s[0]);
            if (s[1] == "1") a[i].isClockWise = true;
            else a[i].isClockWise = false;
        }
        for (int i = 0; i < t; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (a[j].isClockWise) a[j].pos = (a[j].pos + 1) % l;
                else a[j].pos = (a[j].pos + l - 1) % l;
            }
            for (int j = 0; j < n; j++)
            {
                if (a[j].pos - (a[(j + 1) % n].pos + l) % l == 0 && a[j].isClockWise && !a[(j + 1) % n].isClockWise)
                {
                    a[j].isClockWise = false;
                    a[(j + 1) % n].isClockWise = true;
                }
                else if (a[j].pos - (a[(j + 1) % n].pos + l) % l == 1)
                {
                    Ant buf = a[j];
                    a[j] = a[(j + 1) % n];
                    a[(j + 1) % n] = buf;
                }
            }
        }
        for (int i = 0; i < n; i++)
        {
            Console.WriteLine(a[i].pos);
        }
    }
}