Problem:
Score : 300 points

Problem Statement

Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it.

They will share these cards.
First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards.
Here, both Snuke and Raccoon have to take at least one card.

Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively.
They would like to minimize |x-y|.
Find the minimum possible value of |x-y|.

Constraints

2 \leq N \leq 2 \times 10^5

-10^{9} \leq a_i \leq 10^{9}

a_i is an integer.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_{N}

Output

Print the answer.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

1

If Snuke takes four cards from the top, and Raccoon takes the remaining two cards, x=10, y=11, and thus |x-y|=1. This is the minimum possible value.

Sample Input 2

2
10 -10

Sample Output 2

20

Snuke can only take one card from the top, and Raccoon can only take the remaining one card. In this case, x=10, y=-10, and thus |x-y|=20.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"

using ll = long long;

using namespace std;
#define int ll

//0 -> a-1
#define rep(i, a) for (int i = 0; (i) < (int)(a); (i)++)
//a -> b-1
#define reps(i, a, b) for (int i = (int)(a); (i) < (int)(b); (i)++)
//a-1 -> 0
#define rrep(i, a) for (int i = (int)a - 1; (i) >= 0; (i)--)
//a-1 -> b
#define rreps(i, a, b) for (int i = (int)(a)-1; (i) >= (int)(b); (i)--)
#define FOR(x, vec) for (auto& x : vec)
#define MP(a, b) make_pair((a), (b))
#define PB(a) push_back((a))
#define all(v) (v).begin(), (v).end()
//next_permutation(all(v))
#define PERM(v) next_permutation(all(v))
/*sort(all(v));
(v).erase(unique(all(v)), v.end())*/
#define UNIQUE(v) \
	sort(all(v)); \
	(v).erase(unique(all(v)), v.end())
#define CIN(type, x) \
	type x;          \
	cin >> x
#define YES(f) if ((f)) {cout << "YES" << endl;} else {cout << "NO" << endl;}
#define Yes(f) if ((f)) {cout << "Yes" << endl;} else {cout << "No" << endl;}
#define MINV(v) min_element(all(v))
#define MAXV(v) max_element(all(v))
#define MIN3(a, b, c) min(min(a, b), c)
#define MIN4(a, b, c, d) min(MIN3(a, b, c), d)
#define MIN5(a, b, c, d, e) min(MIN4(a, b, c, d), e)
#define MIN6(a, b, c, d, e, f) min(MIN5(a, b, c, d, e), f)
#define MAX3(a, b, c) max(max(a, b), c)
#define MAX4(a, b, c, d) max(MAX3(a, b, c), d)
#define MAX5(a, b, c, d, e) max(MAX4(a, b, c, d), e)
#define MAX6(a, b, c, d, e, f) max(MAX5(a, b, c, d, e), f)
//b is [a, c)
#define RANGE(a, b, c) ((a) <= (b) && (b) < (c))
//c is [a, e) && d is [b, f)
#define RANGE2D(a, b, c, d, e, f) (RANGE((a), (c), (e)) && RANGE((b), (d), (f)))
#define chmin(a, b) a = min(a, (b))
#define chmin3(a, b, c) a = MIN3(a, (b), (c))
#define chmin4(a, b, c, d) a = MIN4(a, (b), (c), (d))
#define chmin5(a, b, c, d, e) a = MIN5(a, (b), (c), (d), (e))
#define chmin6(a, b, c, d, e, f) a = MIN6(a, (b), (c), (d), (e), (f))
#define chmax(a, b) a = max(a, (b))
#define chmax3(a, b, c) a = MAX3(a, (b), (c))
#define chmax4(a, b, c, d) a = MAX4(a, (b), (c), (d))
#define chmax5(a, b, c, d, e) a = MAX5(a, (b), (c), (d), (e))
#define chmax6(a, b, c, d, e, f) a = MAX6(a, (b), (c), (d), (e), (f))
#define fcout cout << fixed << setprecision(12)
#define RS resize
#define CINV(v, N)             \
	do                         \
	{                          \
		v.RS(N);               \
		rep(i, N) cin >> v[i]; \
	} while (0);

#define RCINV(v, N)             \
	do                          \
	{                           \
		v.RS(N);                \
		rrep(i, N) cin >> v[i]; \
	} while (0);
#define MOD 1000000007

template <class T>
inline T GET()
{
	T x;
	cin >> x;
	return x;
}

template<class T, char d='\n'>
struct Vector {
	vector<T> v;
	Vector(int sz) : v(sz) {}
	Vector(int sz, T t) : v(sz, t) {}
	Vector() : v(0) {}
	Vector(vector<T> t) : v(t) {}
	inline auto begin() {return v.begin();}
	inline auto end() {return v.end();}
	inline auto rbegin() {return v.rbegin();}
	inline auto rend() {return v.rend();}
	inline int size() {return v.size();}
	inline auto resize(int sz) {v.resize(sz);}
	inline auto resize(int sz, T t) {v.resize(sz, t);}
	inline auto empty() {v.empty();}
	inline auto reserve(int sz) {v.reserve(sz);}
	inline auto push_back(T t) {v.push_back(t);}
	inline auto pop_back() {v.pop_back();}
	inline auto assign(int sz, const T& t) {v.assign(sz, t);}
	inline auto& front() {return v.front();}
	inline auto& back() {return v.back();}
	inline auto insert(const auto p, const T& t) {v.insert(p, t);}
	inline auto insert(const auto p, auto f, auto l) {v.insert(p, f, l);}
	inline auto erase(const auto p) {v.erase(p);}
	inline auto erase(const auto f, const auto s) {v.erase(f, s);}
	inline auto emplace_back(auto t) {v.emplace_back(t);}
	inline auto swap(auto& f) {v.swap(f.v);}
	inline auto clear() {v.clear();}
	inline auto& operator[](size_t pos) {return v[pos];}
	inline auto operator<(const auto& a) {return v < a.v;}
	inline auto operator>(const auto& a) {return a < (*this);}
	inline auto operator>=(const auto& a) {return !((*this) < a);}
	inline auto operator<=(const auto& a) {return !(a < (*this));}
	inline auto operator==(const auto& a) {return !(a < (*this)) && !((*this) < a);}
	inline friend istream& operator>>(istream& lhs, Vector& rhs) {
		for (auto& x : rhs) lhs >> x;
		return lhs;
	}
	inline friend ostream& operator<<(ostream& lhs, Vector& rhs) {
		for (int i = 0; i + 1 < rhs.size(); i++) lhs << rhs[i] << d;
		if (rhs.size()) lhs << rhs.back();
		return lhs;
	}
	inline auto make_accumulate_sum(T init) {
		Vector<T> res;
		res.push_back(init);
		for (auto x : v) {
			auto y = res.back();
			res.push_back(y + x);
		}
		return res;
	}
};



void init();
void solve();

signed main()
{
	init();
	solve();
}

void init()
{
	int N;
	cin >> N;
	Vector<ll> v(N);
	cin >> v;
	auto s = v.make_accumulate_sum(0);
	ll res = LLONG_MAX;
	reps(i, 1, N) {
		chmin(res, abs(s[i] - (s[N] - s[i])));
	}
	cout << res << endl;
}

void solve()
{
}