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)

/*I MAY NOT GET THE SUCCESS IMMEDIATELY BUT I WILL GET IT FOR SURE*/
#include<bits/stdc++.h>
#define opt std::ios_base::sync_with_stdio(false)
#define I int
#define li		int32_t
#define lli		long long
#define ulli unsigned long long

#define pn			 printf("\n")
#define nl			 cout<<'\n'
#define sf(N)        scanf("%lld",&N)
#define pf(N)        printf("%lld",N)
#define sf1(N1,N2)    scanf("%lld %lld",&N1,&N2)
#define pf1(N1,N2)    printf("%lld %lld",N1,N2)
#define sf2(N1,N2,N3) scanf("%lld %lld %lld",&N1,&N2,&N3)
#define pf2(N1,N2,N3) printf("%lld %lld %lld",N1,N2,N3)

#define sl          cout<<' '
#define ps          printf(" ")

#define rep(i,a,b)	for(i=a;i<b;i++)
#define repr(i,a,b)	for(i=a;i>b;i--)
#define elif		else if
#define mset(a,b)	memset(a,b,sizeof(a))

#define pb		push_back
#define pob		pop_back
#define itr		iterator
#define sz()	size()
#define szof    sizeof
#define lb		lower_bound
#define ub		upper_bound
#define mp		make_pair
#define vlli    vector<lli>
#define plli	pair<lli,lli>
#define vplli	vector<plli >
#define F   	first
#define S		second
#define Dup     erase(unique(V.begin(),V.end()),V.end())

#define bs      binary_search
#define ub      upper_bound
#define lb      lower_bound

#define ALL(V)  V.begin(),V.end()

#define MX1     100000
#define MX2     1000000
#define Inf     1000000000000000
#define mod		1000000007
using namespace std;

lli dp[2*MX1+5];

lli Power(lli a,lli b)
{
    lli result=1;

    while(b)
    {
        if(b%2)
        {
            result=(result*a)%mod;
        }

        b=b>>1;
        a=(a*a)%mod;
    }

 return result;
}

int main()
{
    opt;

    /*#ifndef ONLINE_JUDGE
    freopen("double_squares.txt","r",stdin);
    freopen("double_squares_out.txt","w",stdout);
    #endif // ONLINE_JUDGE*/

    lli N,i;
    cin>>N;

    lli Min=Inf;

    rep(i,1,N+1)
    {
        cin>>dp[i];
    }

    rep(i,1,N+1)
    {
        dp[i]+=dp[i-1];
    }

    rep(i,1,N)
    {
        lli x=dp[i],y=dp[N]-dp[i];

        Min=min(Min,abs(x-y));
    }

    cout<<Min;




 return 0;
}