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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384


int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;} 
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}    //最大公約数
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}     //最小公倍数
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;} //コンビネーション
int nHr(int a,int b){return nCr(a+b-1,b);}       // 重複組み合わせ
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}  //階乗
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;} // a~bまでの階乗
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}   //数字の各位の和
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}  // b進数の各位の和？
int sankaku(int x){return ((1+x)*x)/2;} //三角数 xまでの和

//以下long long ver
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llnHr(long long a,long long b){return llnCr(a+b-1,b);}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long lldsum(long long x){long long r=0;while(x){r+=(x%10);x/=10;}return r;}
long long lldsumb(long long x,long long b){long long r=0;while(x){r+=(x%b);x/=b;}return r;}
long long llsankaku(long long x){return ((1+x)*x)/2;}

//double
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}


typedef pair<int, int> ii;
typedef long long ll;
typedef vector<ll> vll;
typedef pair<ll,ll> pll;
#define forr(i,a,b) ; for(int i=(a); i<(b); i++)
#define clean(arr,val) memset(arr,val,sizeof(arr))
#define forn(i,n) forr(i,0,n)
#define PB push_back
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<pll> vpll;
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
bool is_prime (int x){
  if (x<=1) return false ;
  for (int i=2 ; i*i<=x ; i++){
    if (x%i==0) return false ;
  }
   return true;
}
string alpha = "abcdefghijklmnopqrstuvwxyz" ;
string ALPHA = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" ;
    
bool chmin (int a,int b){
    if ( a > b ){
    a = b;
    return true ;
    }
    return false ;
}
 bool chmax (int a,int b){
    if ( a < b ){
    a = b;
    return true ;
    }
    return false ;
}   
    
 
/*CODE START HERE*/
int main() {
  long long  n;
  cin >> n;
  long long sum = 0;
  vector<long long > v(n);
  forn(i,n){
    cin >> v.at(i);
    sum += v.at(i);
  }
  double bin = sum * 0.5  ;
  double dis = bin;
  long long nsum = 0 ;
  forn ( i,n){
    dis =min ( abs( bin-nsum),dis);
    nsum += v.at(i) ;
  }
  double ans = dis * 2;
  cout << ans << endl;
}