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 <bits/stdc++.h>

using namespace std;

#define INF 1001000100010001000
#define MOD 1000000007
#define EPS 1e-10
#define int long long
#define rep(i, N) for (int i = 0; i < N; i++)
#define Rep(i, N) for (int i = 1; i < N; i++)
#define For(i, a, b) for (int i = (a); i < (b); i++)
#define pb push_back
#define eb emplece_back
#define mp make_pair
#define i_i pair<int, int>
#define vi vector<int>
#define vvi vector<vi >
#define vb vector<bool>
#define vvb vector<vb >
#define vp vector< i_i >
#define Edge pair< i_i, int>
#define all(a) (a).begin(), (a).end()
#define Int(x) int x; scanf("%lld", &x);
#define int2(x, y) int x, y; scanf("%lld %lld", &x, &y);
#define fir first
#define sec second
#define ffir first.first
#define fsec first.second
#define sfir second.first
#define ssec second.second
 
//int dxy[5] = {0, 1, 0, -1, 0};
// assign
 
signed main()
{
    Int(n);
    vi data(n);
    int sum = 0;
    rep(i, n) {
        cin >> data[i];
        sum += data[i];
    }
    int ans = INF;

    int add = 0;
    rep(i, n-1) {
        add += data[i];
        ans = min(ans, abs(add - (sum - add)));
    }
    cout << ans << endl;



    return 0;
}