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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <unordered_set>
#pragma GCC optimize("Ofast")
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int,int> > vpii;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<string> > vvs;
typedef vector<vector<ll> > vvll;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define irep(it, stl) for(auto it = stl.begin(); it != stl.end(); it++)
#define drep(i,n) for(int i = (n) - 1; i >= 0; --i)
#define fin(ans) cout << (ans) << '\n'
#define STI(s) atoi(s.c_str())
#define mp(p,q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Sort(a) sort(a.begin(), a.end())
#define Rort(a) sort(a.rbegin(), a.rend())
#define MATHPI acos(-1)
#define itn int;
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template <class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
struct io{io(){ios::sync_with_stdio(false);cin.tie(0);}};
const int INF = INT_MAX;
const ll LLINF = 1LL<<60;
const ll MOD = 1000000007;
const double EPS = 1e-9;


class Monoid {
public:
    // 単位元
    ll unit;
    
    Monoid() {
        // 単位元
        unit = 0;
    }

    // 演算関数
    ll calc(ll a, ll b) {
        return (a+ b); 
    }
};

class SegmentTree {
public:
    // セグメント木の葉の要素数
    ll n;

    // セグメント木
    vector<ll> tree;

    // モノイド
    Monoid mono;

    SegmentTree(vector<ll> &v) {
        n = 1 << (ll)ceil(log2(v.size()));
        tree = vector<ll>(n << 1);
        for(ll i = 0; i < v.size(); i++) {
            update(i, v[i]);
        }
        for(ll i = v.size(); i < n; i++) {
            update(i, mono.unit);
        }
    }

    // k番目の値(0-indexed)をxに変更
    void update(ll k, ll x) {
        k += n;
        tree[k] = x;
        for(k = k >> 1; k > 0; k >>= 1){
            tree[k] = mono.calc(tree[k << 1 | 0], tree[k << 1 | 1]);
        }
    }

    // [l, r)の最小値(0-indexed)を求める．
    ll query(ll l, ll r) {
        ll res = mono.unit;
        l += n;
        r += n;
        while(l < r) {
            if(l & 1) {
                res = mono.calc(res, tree[l++]);
            }
            if(r & 1) {
                res = mono.calc(res, tree[--r]);
            }
            l >>= 1;
            r >>= 1;
        }
        return res;
    }
    ll operator[](ll k) {
        // st[i]で添字iの要素の値を返す
        if(k - n >= 0 || k < 0) {
            return -INF;
        }
        return tree[tree.size()-n+k];
    }

    void show() {
        ll ret = 2;
        for(ll i = 1; i < 2*n; i++) {
            cout << tree[i] << " ";
            if(i == ret - 1) {
                cout << endl;
                ret <<= 1;
            }
        }
        cout << endl;
    }
};

int main(void) {
    
    ll n;
    cin>>n;
    vll v(n);
    rep(i,n)cin>>v[i];
    SegmentTree st(v);
    ll ans=LLINF;
    rep(i,n-1){
        ans=min(ans,abs(st.query(0,i+1)-st.query(i+1,n)));
    }
    fin(ans);
}