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 <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <iomanip>
#include <math.h>
#include <string.h>
#include <cstdio>
#include <tuple>
using namespace std; using ll = long long; using pll = pair<ll, ll>;
using vl = vector<ll>; using vll = vector<vl>; using vpll = vector<pll>;
//**デバッグ準備**//
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
void debug_out() { cerr << endl; }
template<typename Head, typename... Tail>
void debug_out(Head h, Tail... t) {
	cerr << " " << h;
	if (sizeof...(t) > 0) cerr << " :";
	debug_out(t...);
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> vec) {
	for (size_t i = 0; i < vec.size(); i++)os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
ll ctoi(char c) {
	switch (c) {
	case '0': return 0; case '1': return 1; case '2': return 2;
	case '3': return 3; case '4': return 4; case '5': return 5;
	case '6': return 6; case '7': return 7; case '8': return 8;
	case '9': return 9;  default: return 0;
	}
}
bool pairCompare(const pll firstElof, pll secondElof)
{
	return firstElof.first > secondElof.first;
}
//**Snippetリスト**//
//rep, vin, all, iteLoop//
//bitSearch, bitList, gcdlcm, isPrime, primeFactorize, Npow, divisor, modinv, dfs, bfs, eratos//
//dijkstra, WarshallFloyd, BellmanFord, UnionFind, COM, digitDP ,RMQ//
ll i, j, k, l; ll N, M, K, H, W, L, X, Y, Z, R;
ll MOD = 1000000007; ll INF = 1LL << 62; ll ans = 0; ll z = 0, o = 1;
vl flag, color, D; vll path;
int main() {
	cin >> N;
	vl A(N);
	vl sum(N + 1);
	sum[0] = 0;
	for (i = 0; i < N; i++) {
		cin >> A[i];
		sum[i + 1] = sum[i] + A[i];
	}
	ans = INF;
	for (i = 1; i <= N-1; i++) {
		ans = min(ans, abs(sum[i] - (sum[N] - sum[i])));
	}
	cout << ans;
}