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>
#include <cmath>
#include <numeric>
using namespace std;
#define rep(i,a,b) for(int64_t i=(a); i<(b); ++i) // a ≦ i < b 
#define Rrep(i,a,b) for(int64_t i=(a);i>=(b);--i) // reverse repeat. a から b まで減少．
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend() //逆イテレータ
#define RANGE(a,b,c) (a).begin()+b,(a).begin()+c // コンテナ a の 要素 b から c へのイテレータ
#define MOD 1000000007
#define INF 1000000000
typedef pair<int64_t, int64_t> PII;
typedef vector<int64_t> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef vector<PII> VP;

int main() {
cin.tie(0);
ios::sync_with_stdio(false);

int N; cin >> N;
VI A(N); rep(i, 0, N) cin >> A[i];
int64_t len = accumulate(ALL(A), 0LL);
int64_t r = 0, l =0;

{int i = 0;
while (r < len/2) {
    r += A[i];
    ++i;
}
while (i < N){
    l += A[i];
    ++i;
}
}

cout << r - l << endl;
}

// 境界，出力文字列 チェック
// 可読性優先．高速化次点．
// まずは全探索，分割統治，次にDP
// 制限を見る
// 偶奇，逆から，ソート，出現回数，出現位置，DP， 余事象，包除
// データ構造． 問題の特徴量
// 存在判定：構成方法，入力の特徴
// gcd, lcm ,素因数分解．
// 例外を十分に含む一般化．想像力の限界
// 小さい系から例示
// 始めは過剰に例示・場合分けしてもいい．各場合を確実に対処．
// 自明な例から処理，除外．
// 小数のときは，精度の設定する．doubel 変数に数値を入力するときは 123. とする．
// テストケース作成は数表あり