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>
// #include "ane.cpp"

const int INF  = 1e9;
const long long INFLL = 1e18;
const int NMAX = 200005;
const int MMAX = 100005;
const int KMAX = 1005;
const int MOD  = 1e9 + 7;
using namespace std;

// comment to disable debug functions
#define DEBUG

// frequently used macros

#if __cplusplus >= 201103L
#define ALL(v) begin(v),end(v)
#define SORT(v) sort(begin(v), end(v))
#define FIND(v,x) find(begin(v), end(v), (x))
#else
#define ALL(v) (v).begin(),(v).end()
#define SORT(v) sort(v.begin(), v.end())
#define FIND(v,x) find(v.begin(), v.end(), (x))
#endif

#define MEMNEXT(from, to) do{ memmove((to), (from), sizeof(from)); \
memset((from), 0, sizeof(from)); } while(0)
#ifdef DEBUG
#define DUMP(x) do{ std::cerr << (#x) << ": " << x << std::endl; }while(0)
#else
#define DUMP(x) do{}while(0)
#endif

// frequent used aliases
typedef long long ll;
typedef pair<int, int> p;
typedef pair<ll, int> lp;
typedef pair<int, ll> pl;
typedef pair<ll, ll> llp;
typedef vector<int> vec;
typedef vector<ll> vecll;
typedef vector<vec> mat;
typedef vector<vecll> matll;

// frequently used constants
static const int di[] = {-1, 0, 1, -1, 1, -1, 0, 1};
static const int dj[] = {-1, -1, -1, 0, 0, 1, 1, 1};

// frequently used structs
struct edge{
  int to,cost;
};

// printf for debug
#ifndef DEBUG
void debug(const char* format, ...){}
#else
void debug(const char* format, ...){
  va_list arg;
  va_start(arg, format);
  vprintf(format, arg);
  va_end(arg);
}
#endif

// dump vector
#ifdef DEBUG
#define DUMPV(v, c) do{       \
  printf("%s: ", #v);         \
  for (int i = 0; i < (c); ++i) \
  {                           \
  cout << (v)[i] << " ";      \
  }                           \
  cout << endl;               \
} while(0)
#else
#define DUMPV(v,c)
#endif

// std::fill of multi dimensions
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
  std::fill( (T*)array, (T*)(array+N), val );
}

// binary search
ll BSearch(ll _begin, ll _end, bool (*f)(int)){
  ll mid;
  while(_end - _begin > 1LL) {
  mid = (_begin + _end) / 2LL;
  if(f(mid)) {
    debug("BSearch: f(%d) == true\n", mid);
    _end = mid;
  }
  else
  {
    debug("BSearch: f(%d) == false\n", mid);
    _begin = mid;
  }
  }
  return _end;
}
// binary search
ll BSearchRev(ll _begin, ll _end, bool (*f)(ll)){
  ll mid;
  while(_end - _begin > 1LL) {
  mid = (_begin + _end) / 2LL;
  if(f(mid)) {
    debug("BSearch: f(%d) == true\n", mid);
    _end = mid;
  }
  else
  {
    debug("BSearch: f(%d) == false\n", mid);
    _begin = mid;
  }
  }
  return _end;
}


ll N,M,K,A[NMAX],B,C,D,E;
int dp[NMAX] = {};

string S;
vec v;
std::map<pl, ll> mp;
ll ans = {};

ll solve(int i, ll diff, bool anum, bool bnum){
  if(i >= N){
    if(!anum || !bnum) return INFLL;
    else return abs(diff);
  }
  if (mp.count(pl(i, diff))) return mp[pl(i, diff)];

  return mp[pl(i, diff)] = min(solve(i+1, diff + A[i], true, bnum), solve(i+1, diff - A[i], anum, true));
}
void debug(){
  // output debug information

}
void answer(){
  // output answer
  cout << ans << endl;
}
int main(int argc, char const *argv[])
{
  // operate inputs

  // Fill(dp, -1);
  cin >> N;
  for (int i = 0; i < N; ++i)
  {
    scanf("%lld", &A[i]);
  }
  ans = solve(0, 0, false, false);
  #ifdef DEBUG
  debug();
  #endif
  answer();

  return 0;
}