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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

class Program
{
    static void Main(string[] args)
    {
        var waaaa = new waaaa();
        waaaa.Solve();
    }
}

public class waaaa
{
    const int MOD = 1000000007;
    const long INF = 1 << 62;
    const int MOD2 = 998244353;

    myinout mio = new myinout();
    Math2 mmath = new Math2();

    public void Solve()
    {
        //ここから
        var o = mio.IntM;
        var N=o[0];
        var A=mio.LongM;
        
        long[] l=new long[N+1];
        long[] r=new long[N+1];

        for(int i=0;i<N;++i){
            l[i+1]=l[i]+A[i];
        }
        for(int i=N-1;i>=0;i--){
            r[i]=r[i+1]+A[i];
        }

        long ans=Math.Abs(r[0]-l[0]);
        for(int i=0;i<N;++i){
            ans=Math.Min(ans,Math.Abs(r[i]-l[i]));
        }
        mio.wri(ans);
    }



}

    //b-aで降順

public class Math2
{
    public long[] Fact;  //Nまでの階乗
    public long[] invFact;  //階乗の逆元

    //Nまでの階乗をセットしておく
    public void FactoricalModSet(long N, long MOD = 1000000007)
    {
        Fact = new long[N + 1];
        invFact = new long[N + 1];

        Fact[0] = 1;
        for (long i = 1; i <= N; ++i)
        {
            Fact[i] = Fact[i - 1] * i % MOD;
        }
        invFact[N] = pow(Fact[N], MOD - 2);

        for (long i = N; i > 0; --i)
        {
            invFact[i - 1] = invFact[i] * i % MOD;
        }

    }

    //nCr 引数:n r (mod)
    public long nCr(long n, long r, long MOD = 1000000007)
    {
        return (Fact[n] * invFact[r] % MOD * invFact[n - r] % MOD);
    }

    //階乗(O(logN)) 引数:底x n乗　(MOD)
    public long pow(long x, long n, long MOD = 1000000007)
    {
        long res = 1;
        while (n > 0)
        {
            if ((n & 1) != 0) res = res * x % MOD;
            x = x * x % MOD;
            n >>= 1;
        }
        return res;
    }

    //最大公約数 引数:整数x,y
    public long GCD(long x, long y)
    {
        while (y != 0)
        {
            var r = x % y;
            x = y;
            y = r;
        }
        return x;
    }

    //最小公倍数 引数:整数x,y
    public long LCM(long x, long y)
    {
        return x * y / GCD(x, y);
    }

    //素因数分解 引数:整数N
    public Dictionary<long, int> PrimeFactorization(long N)
    {
        var ret = new Dictionary<long, int>();

        for (long i = 2; i * i <= N; ++i)
        {
            int cnt = 0;
            while (N % i == 0)
            {
                cnt++;
                N /= i;
            }
            if (cnt != 0) ret[i] = cnt;

        }
        if (N >= 2) ret[N] = 1;

        return ret;

    }

    //約数列挙
    public List<long> divisorenumrate(long N)
    {
        var ret = new List<long>();

        for (long i = 1; i * i <= N; ++i)
        {

            if (N % i == 0)
            {
                ret.Add(i);
                ret.Add(N / i);
            }

        }

        return ret;

    }

}

class myinout
{
    //in
    public string Str => Console.ReadLine();

    public string[] StrM => Str.Split(' ');

    public long[] LongM => StrM.Select(long.Parse).ToArray();
    public int[] IntM => StrM.Select(int.Parse).ToArray();
    public char[] CharM => Str.ToArray();
    public long Long1 => LongM[0];
    public int Int1 => IntM[0];

    //out
        public void wri<T>(T x)
    {
        Console.WriteLine(x);
    }

    public void yn(bool x)
    {
        if (x) wri("Yes");
        else wri("No");
    }

}