Problem:
Score : 300 points

Problem Statement

In a flower bed, there are N flowers, numbered 1,2,......,N. Initially, the heights of all flowers are 0.
You are given a sequence h=\{h_1,h_2,h_3,......\} as input. You would like to change the height of Flower k to h_k for all k (1 \leq k \leq N), by repeating the following "watering" operation:

Specify integers l and r. Increase the height of Flower x by 1 for all x such that l \leq x \leq r.

Find the minimum number of watering operations required to satisfy the condition.

Constraints

1 \leq N \leq 100

0 \leq h_i \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
h_1 h_2 h_3 ...... h_N

Output

Print the minimum number of watering operations required to satisfy the condition.

Sample Input 1

4
1 2 2 1

Sample Output 1

2

The minimum number of watering operations required is 2.
One way to achieve it is:

Perform the operation with (l,r)=(1,3).

Perform the operation with (l,r)=(2,4).

Sample Input 2

5
3 1 2 3 1

Sample Output 2

5

Sample Input 3

8
4 23 75 0 23 96 50 100

Sample Output 3

221

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//FIRST THINK THEN CODE.

#include <bits/stdc++.h>
 
 
using namespace std;
 
typedef long long ll;
 
#define MOD 998244353
#define rep(i,a,b) for(ll i=a;i<b;++i)
#define rrep(i,a,b) for(ll i=a;i>b;--i)
#define vi vector<int>
#define vl vector<ll>
#define ld long double
#define vvi vector<vector<int>>
#define vvl vector<vector<long long>>
#define pii pair<int,int>
#define pll pair<long,long>
#define vpii vector<pii>
#define vpll vector<pll>
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define PI 3.1415926535897932384626433832795
#define fix(f,n) fixed<<setprecision(n)<<f
#define all(x) x.begin(),x.end()
#define endl "\n"
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
 
ll newmod(ll a,ll b)
 {
  return ((a%b)+b)%b;
}
 
 
ll powM(ll a,ll b,ll m )
{ 
  a%=m;
  ll ans=1LL;
  while(b)
  {
    if(b&1)ans=ans*a%m;
    a=a*a%m;
    b>>=1;
  }
 
return ans;
}
 
ll poww(ll a,ll b)
{ 
  
  ll ans=1;
  while(b)
  {
    if(b&1)ans=ans*a;
    a=a*a;
    b>>=1;
  }
 
return ans;
}

const ll N=5e5+5;

ll sp[N];

void sieve(){
  for(int i=2;i<N;i++)sp[i]=i;

    for(int i=2;i*i<N;i++){
      if(sp[i]==i){
      for(int j=i*i;j<N;j+=i){
        sp[j]=i;
      }
    }
}

}

ll fac[N],inv[N];

ll p=1e9+7;

void fill(){
  fac[0]=inv[0]=1;

  for(int i=1;i<N;i++){
    fac[i]=fac[i-1]*i%p;
    inv[i]=powM(fac[i],p-2,p);
  }
}

ll c(ll a,ll b,ll p){
if(a<b)return 0;

  return (fac[a]*inv[a-b]%p)*(inv[b])%p;
}


ll dp[3005][3005][2];

// p=998244353;

int main()
{ 

  IOS;


ll n,x=0;
cin>>n;

vl a(n+1);

for(int i=1;i<=n;i++){
  cin>>a[i];
  x+=max(0LL,a[i]-a[i-1]);
}
cout<<x<<endl;

return 0;
 
}