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 Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.math.*;
import java.util.*;
public class Main { 
 
	public static void main(String[] args) { 
		FastScanner I = new FastScanner(); //Input
		OutPut O = new OutPut(); //Output
		long ans = Long.MAX_VALUE;
		int N = I.nextInt();
		long[] pref = new long[N+1];
		for (int i = 1; i <= N; i++) {
			pref[i]=pref[i-1];
			pref[i]+=I.nextLong();
			//O.p(pref[i]+" ");
		}
		for (int cur = 0; cur <= N; cur++) {
			int L = cur;
			int R = N-cur;
			long Rsum = pref[N]-pref[L];
			long curans = Math.abs(Rsum-pref[L]);
			//O.pln(curans);
			ans=Math.min(curans,ans);
		}
		O.pln(ans);
	}
	public static long GCD(long a, long b) {
		if (a==0||b==0) return Math.max(a,b);
		return GCD(Math.min(a, b),Math.max(a, b)%Math.min(a, b));
	}
	public static long FastExp(long base, long exp, long mod) {
		long ans=1;
		while (exp>0) {
			if (exp%2==1) ans*=base;
			exp/=2;
			base*=base;
			base%=mod;
			ans%=mod;
		}
		return ans;
	}
	public static long ModInv(long num,long mod) {return FastExp(num,mod-2,mod);}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());};
	}
	static class OutPut{
		PrintWriter w = new PrintWriter(System.out);
		void pln(int x) {w.println(x);w.flush();}
		void pln(long x) {w.println(x);w.flush();}
		void pln(String x) {w.println(x);w.flush();}
		void pln(char x) {w.println(x);w.flush();}
		void p(int x) {w.print(x);w.flush();}
		void p(long x) {w.print(x);w.flush();}
		void p(String x) {w.print(x);w.flush();}
		void p(char x) {w.print(x);w.flush();}
	}
}