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)

#define TEST_MODE true

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES

#include "bits/stdc++.h"
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i, a, b) for (int i = (a); i < (b); ++i)
#define rrep(i, n) for (int i = (n)-1; i >= 0; --i)
#define rrep2(i, a, b) for (int i = (a)-1; i >= b; --i)
#define range(i, a, b, c) for (int i = a;             \
                               c > 0 ? i < b : i > b; \
                               i += c)
#define chmax(a, b) (a = (a) < (b) ? (b) : (a))
#define chmin(a, b) (a = (a) > (b) ? (b) : (a))
using ll = long long;
using ull = unsigned long long;
using ld = long double;
#define all(a) begin(a), end(a)
#define ifnot(a) if (not(a))
#define int long long

#ifdef LOCAL_ENV

#if TEST_MODE == true
const bool test = true;
#define dump(x) cerr << #x << " = " << (x) << endl
#else
const bool test = false;
#define dump(x) 
#endif

#else
const bool test = false;
#define dump(x) 
#endif

int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int INF = (int)1 << 60;
const ll INFL = (ll)1 << 60;
ll mod_n = (int)1e9 + 7;
const double eps = 1e-10;
typedef long double Real;
// return -1, 0, 1
int sgn(const Real &r) { return (r > eps) - (r < -eps); }
int sgn(const Real &a, const Real &b) { return sgn(a - b); }

//.....................
const int MAX = (int)2e5 + 5;

vector<string> split(const string &str, char sep)
{
    vector<string> v;
    stringstream ss(str);
    string buffer;
    while (getline(ss, buffer, sep))
    {
        v.push_back(buffer);
    }
    return v;
}

template <class InputIterator>
int sum(InputIterator begin, InputIterator end)
{
    return accumulate(begin, end, 0ll);
}

void solve();

class Solver {
    public:
    int N;
    int a[MAX];
    void solve() {
        cin >> N;
        rep(i, N) cin >> a[i];
        int sum_v1 = a[0];
        int sum_v2 = sum(a, a + N) - a[0];
        int res = abs(sum_v1 - sum_v2);
        rep2(i, 1, N) {
            res = min(res, abs(sum_v1 - sum_v2));
            sum_v1 += a[i];
            sum_v2 -= a[i];
        }
        cout << res << endl;
    }
};

signed main()
{
    cout << fixed << setprecision(20);
    auto ptr = new Solver();
    ptr->solve();
    delete ptr;
    // while (true) {    
    //     auto ptr = new Solver();
    //     ptr->solve();
    //     delete ptr;
    // }
    return 0;
}

// class Mycin {
// 	bool flag = true;
// public:
// 	Mycin& operator >> (int& a) {flag = scanf("%lld", &a) != EOF; return *this;}
// 	Mycin& operator >> (char& a) {flag = scanf("%c", &a) != EOF; return *this;}
// 	Mycin& operator >> (string& s) {flag = (bool)(cin >> s); return *this;}
// 	operator bool() {return flag;}
// } mycin;
 
// class Mycout {
// public:
// 	Mycout& operator << (const int& a) {printf("%lld", a); return *this;}
// 	Mycout& operator << (const char c) {printf("%c", c); return *this;}
// 	Mycout& operator << (const string& s) {printf("%s", s.c_str()); return *this;}
// } mycout;
 
// #define cin mycin
// #define cout mycout
// #define endl '\n'