Problem:
Score : 700 points

Problem Statement

You are given N integers A_1, A_2, ..., A_N.

Consider the sums of all non-empty subsequences of A. There are 2^N - 1 such sums, an odd number.

Let the list of these sums in non-decreasing order be S_1, S_2, ..., S_{2^N - 1}.

Find the median of this list, S_{2^{N-1}}.

Constraints

1 \leq N \leq 2000

1 \leq A_i \leq 2000

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the median of the sorted list of the sums of all non-empty subsequences of A.

Sample Input 1

3
1 2 1

Sample Output 1

2

In this case, S = (1, 1, 2, 2, 3, 3, 4). Its median is S_4 = 2.

Sample Input 2

1
58

Sample Output 2

58

In this case, S = (58).

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define INF 1LL << 60
#define MOD 1000000007ll
#define EPS 1e-10
#define REP(i, m) for (long long i = 0; i < m; i++)
#define FOR(i, n, m) for (long long i = n; i < m; i++)
#define pb push_back
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
typedef long double ld;

int main(){
  bitset<4000000> x = 1;
  int n;
  cin >> n;
  int sum = 0;
  REP(i, n){
    int a;
    cin >> a;
    sum += a;
    x |= x << a;
  }
  int k = sum/2;
  while(true){
    if(x[k]){
      cout << k << endl;
      return 0;
    }
    k++;
  }
	return 0;
}