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: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = double;
using ld = long double;
template<typename T> using V = vector<T>;
template<typename T> using VV = vector<vector<T>>;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define all(v) (v).begin(),(v).end()
#define siz(v) (ll)(v).size()
#define rep(i,a,n) for(ll i=a;i<(ll)(n);i++)
#define repr(i,a,n) for(ll i=n-1;(ll)a<=i;i--)
#define lb lower_bound
#define ub upper_bound
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
const ll mod = 1000000007;
const ll INF = 1000000099;
const ll LINF = (ll)(1e18 +99);
vector<ll> dx={-1,1,0,0},dy={0,0,-1,1};
template<typename T,typename U> void chmin(T& t, const U& u){if(t>u)t=u;}
template<typename T,typename U> void chmax(T& t, const U& u){if(t<u)t=u;}
template<typename T>T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<typename T>T mpow(T a, T n) {
  T res = 1;
  for(;n;n>>=1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}



signed main(){
  cin.tie(0);ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(20);
  ll n;cin>>n;
  vector<ll> v(n),sum(n+1,0);
  rep(i,0,n){
     cin>>v[i];
     sum[i+1]=sum[i]+v[i];
  }

   ll ans=LINF;
  rep(i,0,n){
     ans=min(ans,abs(2*sum[i+1]-sum[n]));
  }
  cout<<ans<<endl;
}
//( ・ __ ・ ) KEEP BEING ORGANIZED
//CHECK overflow,vector_size,what to output?