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 <vector>
#include <algorithm>
#include <iostream>
#include <cassert>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <ctime>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <cassert>
#include <stack> 

#define INFLL 2000000000000000000
#define INF 2000000000
#define MOD 1000000007
#define PI acos(-1.0)
#define MAXLEN 29

typedef std::pair <int, int> pii;
typedef long long ll;
typedef std::vector <ll> vll;

using namespace std;

int arr[2000];
int n;
vector <int> vec;

int main() {
	//freopen("greedy.in", "r", stdin);
	//freopen("greedy.out", "w", stdout);	
	int sum = 0;	
	cin >> n;
	int maxi = -INF;
	for (int i = 0; i < n; i++) {
		cin >> arr[i];
		maxi = max(maxi, arr[i]);
	}
	sort(arr, arr + n);
	for (int i = 0; i < n; i++) sum += arr[i];
	if (n < 17) {
		for (int mask = 1; mask < (1 << n); mask++) {
			int ss = 0;
			for (int i = 0; i < n; i++) {
				if (mask & (1 << i)) ss += arr[i];
			}
			vec.push_back(ss);
		}
		sort(vec.begin(), vec.end());
		cout << vec[vec.size() / 2] << endl;
		return 0;
	}
	cout << maxi << endl;
	return 0;
}