Problem:
Score: 200 points

Problem Statement

Takahashi, who works at DISCO, is standing before an iron bar.
The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.

Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length.
However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:

Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).

Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.

Find the minimum amount of money needed before cutting the bar into two parts with the same length.

Constraints

2 \leq N \leq 200000

1 \leq A_i \leq 2020202020

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_N

Output

Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.

Sample Input 1

3
2 4 3

Sample Output 1

3

The initial lengths of the sections are [2, 4, 3] (in millimeters). Takahashi can cut the bar equally after doing the following operations for 3 yen:

Shrink the second section from the left. The lengths of the sections are now [2, 3, 3].

Shrink the first section from the left. The lengths of the sections are now [1, 3, 3].

Shrink the second section from the left. The lengths of the sections are now [1, 2, 3], and we can cut the bar at the second notch from the left into two parts of length 3 each.

Sample Input 2

12
100 104 102 105 103 103 101 105 104 102 104 101

Sample Output 2

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define double long double
#define fo(a,b) for(int a=0;a<b;a++)
#define Sort(a) sort(a.begin(),a.end())
#define rev(a) reverse(a.begin(),a.end())
#define fi first
#define se second
#define co(a) cout<<a<<endl
#define sz size()
#define bgn begin()
#define en end()
#define pb(a) push_back(a)
#define pp() pop_back()
#define V vector
#define P pair
#define V2(a,b,c) V<V<int>> a(b,V<int>(c))
#define V2a(a,b,c,d) V<V<int>> a(b,V<int>(c,d))
#define incin(a) int a; cin>>a
#define yuko(a) setprecision(a)
#define uni(a) a.erase(unique(a.begin(),a.end()),a.end())
#define pri priority_queue
#define syo <int,V<int>
#define ff fi.fi
#define fs fi.se
#define sf se.fi
#define ss se.se
#define all(a)  (a).begin(),(a).end()
#define Pi P<int,int>
#define elif else if
//#define min min<int>
//#define max max<int>
int low(V<int> a,int b){
  decltype(a)::iterator c=lower_bound(a.begin(),a.end(),b);
  int d=c-a.bgn;
  return d;
}
int upp(V<int> a,int b){
  decltype(a)::iterator c=upper_bound(a.begin(),a.end(),b);
  int d=c-a.bgn;
  return d;
}
template<class T>
 void cou(vector<vector<T>> a){
  int b=a.size();
  int c=a[0].size();
  fo(i,b){
    fo(j,c){
      cout<<a[i][j];
      if(j==c-1)
        cout<<endl;
      else
        cout<<' ';
    }
  }
}
/*template<>
 void cou(vector<vector<char>> a){
  int b=a.size();
  int c=a[0].size();
  fo(i,b){
    fo(j,c){
      cout<<a[i][j];
      if(j==c-1)
        cout<<endl;
      else
        cout<<' ';
    }
  }
}*/
 int wari(int a,int b) {
   if(a%b==0)
     return a/b;
   else
     return a/b+1;
 }
 int keta(int a){
  double b=a;
  b=log10(b);
  int c=b;
  return c+1;
 }
 int souwa(int a){
  return a*(a+1)/2;
 }
 /*int lcm(int a,int b){
  int d=a,e=b,f;
  if(a<b)
    swap(a,b);
  int c,m=1;
  while(m){
    c=a%b;
    if(c==0){
       f=b;
      m--;
    }
    else{
      a=b;
      b=c;
 }
  }
  return d*e/f;
    } */
int gcm(int a,int b){
  if(a%b==0)
    return b;
  return gcm(b,a%b);
}
 bool prime(int a){
   if(a<2)
     return false;
   else if(a==2)
     return true;
   else if(a%2==0)
     return false;
   double b=sqrt(a);
    for(int i=3;i<=b;i+=2){
      if(a%i==0){
        return false;
      }
    }
   return true;
 }

struct Union{
  vector<int> par;
  Union(int a){
  par= vector<int>(a,-1);
  }
  int find(int a){
    if(par[a]<0)
      return a;
    else
      return par[a]=find(par[a]);
  }
  bool same(int a,int b){
    return find(a)==find(b);
  }
  int Size(int a){
    return -par[find(a)];
  }
  void unite(int a,int b){
    a=find(a);
    b=find(b);
    if(a==b)
      return;
    if(Size(b)>Size(a))
      swap<int>(a,b);
    par[a]+=par[b];
    par[b]=a;
  }
};
int ketas(int a){
  string b=to_string(a);
  int c=0;
  fo(i,keta(a)){
    c+=b[i]-'0';
  }
  return c;
}
/*int gcm(int a,int b){
    if(b==0)
        return a;
    return gcm(b,a%b);
}*/
int lcm(int a,int b){
    return a/gcm(a,b)*b;
}
/*struct aa{
  vector<int> gt;
  aa(int n){
  gt= vector<int>(n, 1);
  }
  void c(V<int> d,int b){
  if(d[b]==0){
    gt[d[b]-1]++;
    gt[gt.sz-1]++;
  }
  else{
    gt[d[b]-1]++;
    c(d,d[d[b]]-1);
  }
}
void cok(int a){
  cout<<gt[a-1]<<endl;
  fo(i,a-1)
  cout<<gt[i]<<endl;
}
};
*/
/*struct dfs(){

}*/
bool fe(int a,int b){
  a%=10;
  b%=10;
  if(a==0)
    a=10;
  if(b==0)
    b=10;
  if(a>b)
    return true;
  else
    return false;
}
int INF=1000000007;
struct edge{int s,t,d; };
V<int> mojisyu(string a){
  V<int> b(26,0);
  fo(i,a.sz){
    b[a[i]-'a']++;
  }
  return b;
}
int wa2(int a){
  if(a%2==1)
    return a/2;
  return a/2-1;
}
/*signed main(){
  int a,b,c;
  cin>>a>>b>>c;
  V<V<edge>> d(a);
  fo(i,b){
    edge e;
    cin>>e.s>>e.t>>e.d;
    d[e.s].pb(e);
  }
  V<int> e(a,INF);
  e[c]=0;
  priority_queue<P<int,int>,V<P<int,int>>,greater<P<int,int>>> f;
  f.push({0,c});
  int h=INF;
  while(!f.empty()){
    P<int,int> g;
    g=f.top();
    f.pop();
    int v = g.second, i = g.first;
    for(edge l : d[v]){
      if(e[l.t] > i + l.d){
        e[l.t] = i + l.d;
        f.push({i+l.d , l.t});
      }
    }
  }
  fo(i,a){
    if(e[i]==INF)
      cout<<"INF"<<endl;
    else
      cout<<e[i]<<endl;
  }
}
?*/
signed main(){
   int N,j=0;
   cin>>N;
   V<int> A(N);
   double a=0,x=0,y;
   fo(i,N) cin>>A[i];
   fo(i,N) a+=A[i];
   while(x<a/2){
    x+=A[j];
    j++;
   }
   if(x==a/2) co(0);
   else{
  y=x-A[j-1];
  if(a/2-y>x-a/2) co(x*2-a);
  else co(a-y*2);
  
}
}