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++11 (GCC 4.8.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <vector>
#include <valarray>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <bitset>
#include <utility>
#include <numeric>
#include <algorithm>
#include <functional>
#include <complex>
#include <string>
#include <sstream>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cmath>

using namespace std;

#define all(c) c.begin(),c.end()
#define repeat(i,n) for(int i=0;i<static_cast<int>(n);i++)
#define debug(x) #x << "=" << (x)
#define dump(x) cerr << debug(x) << " (L:" << __LINE__ << ")"<< endl

typedef long long ll;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef vector<long> vl;
typedef vector<vector<long> > vvl;
typedef vector<string> vs;

template<typename T>
ostream& operator<<(ostream& os, const vector<T>& vec){
    os << "[";
    for(int i = 0; i < vec.size(); i++){
        os << vec[i] << ",";
    }
    os << "]";
    return os;
}

template<typename T>
T input(){
    T t;
    cin >> t;
    return t;
}

template<typename T>
vector<T> input(const int N){
    vector<T> v(N);
    repeat(i,N) cin >> v[i];
    return v;
}

long long gcd(long long a, long long b){
    if(b == 0){
        return a;
    }
    return gcd(b, a % b);
}

long long lcm(long long a, long long b){
    return (a / gcd(a, b)) * b;
}

long long mul(const long long& a, const long long& b, const long long& mod) {
    return ((a % mod) * (b % mod)) % mod;
}

long long power(const long long& x, const long long& y, const long long& mod) {
    if (y == 0) {
        return 1;
    } else if (y == 1) {
        return x % mod;
    } else {
        long long value = power(x, y / 2, mod);
        if (y % 2 == 0) {
            return mul(value, value, mod);
        } else {
            return mul(value, value, mod) * x % mod;
        }
    }
}

long long div(const long long& a, const long long& b, const long long& mod) {
    return mul(a, power(b, mod - 2, mod), mod);
}

map<long long, long long> factorials;
long long factorial(const long long& n, const long long& mod) {
    if (n == 0 || n == 1) {
        return 1;
    }
    if (factorials[n] != 0) {
        return factorials[n];
    }
    factorials[n] = n * factorial(n - 1, mod) % mod;
    return factorials[n] % mod;
}

long long combination(const long long& n, const long long& x, const long long& mod) {
    long long numerator = 1;
    long long denominator = 1;
    repeat(i, x) {
        numerator *= (n - i) % mod;
        numerator %= mod;
        denominator *= (i + 1) % mod;
        denominator %= mod;
    }
    return div(numerator, denominator, mod);
}

int main(){
    int N;
    cin >> N;

    vl numbers = input<long>(N);
    vl sumsForward(N, 0);
    vl sumsBackward(N, 0);

    repeat (i, N) {
        if (i == 0) {
            sumsForward[i] = numbers[i];
            sumsBackward[N - 1 - i] = numbers[N - 1 - i];
        } else {
            sumsForward[i] = numbers[i] + sumsForward[i - 1];
            sumsBackward[N - 1 - i] = numbers[N - 1 - i] + sumsBackward[N - i];
        }
    }
    long minDiff = abs(sumsForward[0] - sumsBackward[1]);
    repeat (i, N - 1) {
        long diff = abs(sumsForward[i] - sumsBackward[i + 1]);
        minDiff = min(diff, minDiff);
    }
    cout << minDiff << endl;
    return 0;
}