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++14 (GCC 5.4.1) (judge verdict: Accepted)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author majk
 */

#ifndef MAJK_LIB
#define MAJK_LIB

#include <vector>
#include <stack>
#include <iostream>
#include <unordered_map>
#include <map>
#include <iomanip>
#include <set>
#include <functional>
#include <fstream>
#include <algorithm>
#include <cassert>
#include <cmath>
#include <string>
#include <sstream>
#include <queue>
#include <bitset>
using namespace std;

#define x first
#define y second
typedef std::pair<int,int> pii; typedef long long ll; typedef unsigned long long ull; typedef unsigned int ui; typedef pair<ui,ui> puu;

template <typename T, typename U> std::istream&operator>>(std::istream&i, pair<T,U>&p) {i >> p.x >> p.y; return i;}
template<typename T>std::istream&operator>>(std::istream&i,vector<T>&t) {for(auto&v:t){i>>v;}return i;}
template<typename T>std::ostream&operator<<(std::ostream&o,const vector<T>&t) {if(t.empty())o<<'\n';for(size_t i=0;i<t.size();++i){o<<t[i]<<" \n"[i == t.size()-1];}return o;}
template <typename T, typename U> std::ostream&operator<<(std::ostream&o, pair<T,U>&p) {o << p.x << ' ' << p.y; return o;}
template<typename T> using minheap = priority_queue<T, vector<T>, greater<T>>;
template<typename T> using maxheap = priority_queue<T, vector<T>, less<T>>;
auto fraclt = [](const pii&a,const pii&b) { return (ll)a.x * b.y < (ll)b.x * a.y; };
struct cmpfrac { bool operator()(const pii&a,const pii&b)const { return (ll)a.x * b.y < (ll)b.x * a.y; }};
template <typename T> bool in(T a, T b, T c) { return a <= b && b < c; }
ui logceil(int x) { return 8*sizeof(int)-__builtin_clz(x); }

namespace std {
    template<typename T,typename U>struct hash<pair<T,U>>{hash<T>t;hash<U>u;size_t operator()(const pair<T,U>&p)const{return t(p.x)^(u(p.y)<<7);}};
}
template<typename T,typename F>T bsh(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){l=m+1;r=m;}else{h=m-1;}}return r;}
template<typename F> double bshd(double l,double h,const F&f,double p=1e-9){ui r=3+(ui)log2((h-l)/p);while(r--){double m=(l+h)/2;if(f(m)){l=m;}else{h=m;}}return (l+h)/2;}
template<typename T,typename F>T bsl(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){h=m-1;r=m;}else{l=m+1;}}return r;}
template<typename F> double bsld(double l,double h,const F&f,double p=1e-9){ui r=3+(ui)log2((h-l)/p);while(r--){double m=(l+h)/2;if(f(m)){h=m;}else{l=m;}}return (l+h)/2;}
template<typename T> T gcd(T a,T b) { if (a<b) swap(a,b); return b?gcd(b,a%b):a; }

template<typename T>class vector2:public vector<vector<T>>{public:vector2(){} vector2(size_t a,size_t b,T t=T()):vector<vector<T>>(a,vector<T>(b,t)){}};
template<typename T>class vector3:public vector<vector2<T>>{public:vector3(){} vector3(size_t a,size_t b,size_t c,T t=T()):vector<vector2<T>>(a,vector2<T>(b,c,t)){}};
template<typename T>class vector4:public vector<vector3<T>>{public:vector4(){} vector4(size_t a,size_t b,size_t c,size_t d,T t=T()):vector<vector3<T>>(a,vector3<T>(b,c,d,t)){}};
template<typename T>class vector5:public vector<vector4<T>>{public:vector5(){} vector5(size_t a,size_t b,size_t c,size_t d,size_t e,T t=T()):vector<vector4<T>>(a,vector4<T>(b,c,d,e,t)){}};
template <typename T> struct bounded_priority_queue {
	inline bounded_priority_queue(ui X) : A(X), B(0) {}
	inline void push(ui L, T V) { B = max(B, L); A[L].push(V); }
	inline const T &top() const { return A[B].front(); }
	inline void pop() { A[B].pop(); while (B > 0 && A[B].empty()) --B; }
	inline bool empty() const { return A[B].empty(); }
	inline void clear() { B = 0; for (auto &a: A) a = queue<T>(); }
private:
	vector<queue<T>> A; ui B;
};


#endif
// #include "../l/mod.h"

class A {
public:
    void solve(istream& cin, ostream& cout) {
        int N; cin >> N;
        vector<int> A(N); cin >> A;
        int ans = 1, b = 0;
        for (int i = 1; i < N; ++i) {
            if ((A[b] < A[i-1] && A[i-1] > A[i]) || (A[b] > A[i-1] && A[i-1] < A[i])) {
                ++ans;
                b = i;
            }
        }
        cout << ans << endl;
    }
};


int main() {
	ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	A solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
    return 0;
}