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 <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <numeric>
#include <utility>
#include <sstream>
#include <cassert>
template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
using namespace std;
#define int long long
#define ll long long
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define P pair<ll, ll>
#define sz(x) (ll)x.size()
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define VE vector<ll>
#define COUT(x) cout<<(x)<<endl
#define MA map<ll,ll>
#define SE set<ll>
#define PQ priority_queue<ll>
#define PQR priority_queue<ll,VE,greater<ll>>
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define EPS (1e-10)
#define pb push_back
long long MOD = 1000000007;
//long long MOD = 998244353;
const long long INF = 1LL << 60;
const long double PI = acos(-1.0);
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x%MOD + MOD) % MOD) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const mint a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint a) const {
		mint res(*this);
		return res *= a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}
	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const {
		mint res(*this);
		return res /= a;
	}
	bool operator==(const mint a) const {
		return x == a.x;
	}
	bool operator!=(const mint a) const {
		return x != a.x;
	}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
struct Sieve {
	int n;
	vector<int> f, primes;

	Sieve(int n = 1) :n(n), f(n + 1) {
		f[0] = f[1] = -1;
		for (ll i = 2; i <= n; ++i) {
			if (f[i]) continue;
			primes.push_back(i);
			f[i] = i;
			for (ll j = i * i; j <= n; j += i) {
				if (!f[j]) f[j] = i;
			}
		}
	}

	bool isPrime(int x) { return f[x] == x; }

	vector<int> factorList(int x) {
		vector<int> res;
		while (x != 1) {
			res.push_back(f[x]);
			x /= f[x];
		}
		return res;
	}

	vector<P> factor(int x) {
		vector<int> fl = factorList(x);
		if (fl.size() == 0) return {};
		vector<P> res(1, P(fl[0], 0));
		for (int p : fl) {
			if (res.back().first == p) {
				res.back().second++;
			}
			else {
				res.emplace_back(p, 1);
			}
		}
		return res;
	}
};
template<class t> t gcd(t a, t b) {
	return b != 0 ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}
bool prime(ll n) {
	for (ll i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
map<ll, ll> prime_factor(ll n) {
	map<ll, ll> ret;
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret[i]++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}
ll modinv(ll a, ll m) {
	ll b = m, u = 1, v = 0;
	while (b) {
		ll t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
vector<pair<char, int>> RunLength(string s) {
	if (s.size() == 0)return {};
	vector<pair<char, int>>res(1, pair<char, int>(s[0], 0));
	for (char p : s) {
		if (res.back().first == p) {
			res.back().second++;
		}
		else {
			res.emplace_back(p, 1);
		}
	}
	return res;
}
// Digit Count
int GetDigit(int num) {
	return log10(num) + 1;
}
// bit calculation[how many "1"] (= __builtin_popcount())
int bit_count(int n) {
	int cnt = 0;
	while (n > 0) {
		if (n % 2 == 1)cnt++;
		n /= 2;
	}
	return cnt;
}
vector<long long> enum_divisors(long long N) {
	vector<long long> res;
	for (long long i = 1; i * i <= N; ++i) {
		if (N % i == 0) {
			res.push_back(i);
			if (N / i != i) res.push_back(N / i);
		}
	}
	sort(res.begin(), res.end());
	return res;
}
const ll dx[4] = { 1,0,-1,0 };
const ll dy[4] = { 0,1,0,-1 };
//struct edge { ll to, cost; };
typedef long double ld;
using Graph = vector<vector<int>>;
class UnionFind {
public:
	vector <ll> par; // 各元の親を表す配列
	vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1) {
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.resize(sz_, 1);
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}

	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}
};
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		//assert(n < mod);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
};
struct edge { int to, cost; };


signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	//cout << fixed << setprecision(20);
	//combination com(200010);
	int n; cin >> n;
	VE a(n); rep(i, n)cin >> a[i];
	VE sum(n + 1);
	rep(i, n) {
		sum[i + 1] = sum[i] + a[i];
	}
	int ans = INF;
	for (int i = 1; i < n; i++) {
		chmin(ans, abs(sum[i] - (sum[n] - sum[i])));
	}
	cout << ans << endl;
	return 0;
}