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++ (GCC 9.2.1) (judge verdict: Accepted)

    /**   Created by: Marcos Portales
          Codeforces User: marcosportales
          Country: Cuba
          Copyright                    */
#include<bits/stdc++.h>
using namespace std;
/// Pragmas:
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline") //Optimization flags
#pragma GCC option("arch=native","tune=native","no-zero-upper") //Enable AVX
#pragma GCC target("avx2") //Enable AVX
/// Macros:
#define int long long
#define f first
#define s second
#define db(x) cerr << #x << ": " << (x) << '\n';
#define pb push_back
#define lb lower_bound
#define up upper_bound
#define all(x) x.begin() , x.end()
#define rall(x) x.rbegin() , x.rend()
#define enl '\n'
#define vi vector<int>
#define sz(a) int((a).size())
#define rep(i,n) for(int i=0;i<n;i++)
#define rep2(i,x,y) for(int i=(x);i<(y);i++)
typedef pair<int,int> ii;
typedef long double ld;
typedef unsigned long long ull;
/// Constraints:
const int maxn = 400010;
const int mod = 1000000007;
const ld eps = 1e-9;
const int inf = ((1ll<<31ll)-1ll);
const int INF = 2000000000000000000ll;
const ld pi = acos(-1);
/// Prime Numbers:
// 2, 173, 179, 181, 197, 311, 331, 737, 1009, 2011, 2027, 3079, 4001, 10037, 10079, 20011, 20089;
// 100003, 100019, 100043, 200003, 200017, 1000003, 1000033, 1000037, 1000081;
// 2000003, 2000029, 2000039, 1000000007, 1000000021, 2000000099;
/// Functions:
#define lg2(x) 31 - __builtin_clz(x)
#define lgx(x,b) ( log(x) / log(b) )
/// Red-Black Tree Test.insert(x)late ---------------------------------
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree < long long ,  null_type ,  less<long long> ,  rb_tree_tag ,  tree_order_statistics_node_update > ordered_set;
/// Quick Pow------------------------------------------------
int qpow(int b,int e){
    if( !e ) return 1;
    if( e & 1 ) return qpow(b,e-1) * b % mod;
    int pwur = qpow(b,e>>1);
    return (pwur * pwur) % mod;
}
int modinv(int x){
    return qpow(x,mod-2);
}
/// My Code -------------------------------------------------

int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cout.setf(ios::fixed); cout.precision(0);
    srand(time(NULL));
    //freopen("a.in","r",stdin);
    //freopen("a.in","w",stdout);
    int n;
    cin>>n;
    vi v(n),S(n);
    rep(i,n){
        cin>>v[i];
    }
    S[0]=v[0];
    rep2(i,1,n){
        S[i]=S[i-1]+v[i];
    }
    int ans=inf;
    rep(i,n-1){
        ans=min(ans,abs(S[i]-(S[n-1]-S[i])));
    }
    cout<<ans;

    return 0;
}