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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;  
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
#define rep(i,n) for(ll i=0;i<(n);i++)  
#define pii pair<int,int>
#define piii pair<int,pii>
#define mp make_pair
#define pb push_back  
#define ALL(a) (a).begin(),(a).end()
#define FST first
#define SEC second  
const int INF = (INT_MAX/2);
const ll LINF = (LLONG_MAX/2);
const double eps = 1e-14;
const double PI = M_PI;  
#define DEB cout<<"!"<<endl
#define SHOW(a,b) cout<<(a)<<" "<<(b)<<endl
#define SHOWARRAY(ar,i,j) REP(a,i)REP(b,j)cout<<ar[a][b]<<((b==j-1)?((a==i-1)?("\n\n"):("\n")):(" "))
  
#define DIV 1000000007

typedef vector<ll> Array;
typedef vector<Array> matrix;

int main(){
  Array v(200000);
  int n; cin >> n;
  ll sum = 0;
  rep(i,n){
    cin >> v[i];
    sum += v[i];
  }
  
  ll ans = INF;
  
  ll now = 0;
  
  for(int i = 0; i < n - 1; i++){
    now += v[i];
    ans = min(ans,abs(now - (sum-now)));
  }
  cout << ans << endl;
}