Problem:
Score : 1400 points

Problem Statement

Snuke has a sequence of N integers x_1,x_2,\cdots,x_N.
Initially, all the elements are 0.

He can do the following two kinds of operations any number of times in any order:

Operation 1: choose an integer k (1 \leq k \leq N) and a non-decreasing sequence of k non-negative integers c_1,c_2,\cdots,c_k.
Then, for each i (1 \leq i \leq k), replace x_i with x_i+c_i.

Operation 2: choose an integer k (1 \leq k \leq N) and a non-increasing sequence of k non-negative integers c_1,c_2,\cdots,c_k.
Then, for each i (1 \leq i \leq k), replace x_{N-k+i} with x_{N-k+i}+c_i.

His objective is to have x_i=A_i for all i.
Find the minimum number of operations required to achieve it.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N

Output

Print the minimum number of operations required to achieve Snuke's objective.

Sample Input 1

5
1 2 1 2 1

Sample Output 1

3

One way to achieve the objective in three operations is shown below.
The objective cannot be achieved in less than three operations.

Do Operation 1 and choose k=2,c=(1,2). Now we have x=(1,2,0,0,0).

Do Operation 1 and choose k=3,c=(0,0,1). Now we have x=(1,2,1,0,0).

Do Operation 2 and choose k=2,c=(2,1). Now we have x=(1,2,1,2,1).

Sample Input 2

5
2 1 2 1 2

Sample Output 2

2

Sample Input 3

15
541962451 761940280 182215520 378290929 211514670 802103642 28942109 641621418 380343684 526398645 81993818 14709769 139483158 444795625 40343083

Sample Output 3

7

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 <fstream>
#include <unordered_map>
#include <unordered_set>
#include <map>
#include <iomanip>
#include <set>
#include <functional>
#include <algorithm>
#include <numeric>
#include <cassert>
#include <cmath>
#include <string>
#include <queue>
#include <array>
#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, typename U> std::ostream&operator<<(std::ostream&o, const pair<T,U>&p) {o << p.x << ' ' << p.y; return o;}
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> using minheap = priority_queue<T, vector<T>, greater<T>>;
template<typename T> using maxheap = priority_queue<T, vector<T>, less<T>>;
ui logceil(ll x) { return x?8*sizeof(ll)-__builtin_clzll(x):0; }

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)){}};


#endif

class EPrefixSuffixAddition {
public:
    void solve(istream& cin, ostream& cout) {
        int N; cin >> N;
        vector<int> A(N+2, 0);
        for (int i = 0; i < N; ++i) cin >> A[i+1];
        int x = 0, y = 0;
        vector<pair<int,pii>> X{{0,{0,0}}};
        for (int i = 1; i <= N+1; ++i) {
            x += A[i-1] > A[i];
            y += A[i-1] < A[i];
            auto cost = [&](int cur) {
                int ans = 2*N;
                for (auto &x: X) {
                    ans = min(ans, x.x + (x.y.x > cur) + (A[i-1]-x.y.x < A[i]-cur));
                }
                return ans;
            };
            int hi = cost(0);
            int lo = cost(A[i]);
            vector<pair<int,pii>> Y;
            for (int a = lo; a <= hi; ++a) {
                int b = bsl(0, A[i], [&](int x) { return cost(x) <= a; });
                int c = bsh(0, A[i], [&](int x) { return cost(x) >= a; });
                if (b <= c) Y.push_back({a, {b,c}});
            }
            swap(Y,X);
        }
        cout << X[0].x << '\n';
    }
};


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