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: Accepted)

#include <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>
#include <bitset>
#include <map>
#include <set>
#include <tuple>
#include <string.h>
#include <math.h>
#include <random>
#include <functional>
#include <assert.h>

using namespace std;

using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;

bitset<4000001> sums;

int main()
{
    sums[0] = 1;

    int n;
    scanf("%d", &n);

    int sum = 0;

    for (int i = 0; i < n; i++)
    {
        int k;
        scanf("%d", &k);
        sum += k;
        
        sums |= sums << k;
    }

    for (int i = (sum+1) / 2;; i++)
    {
        if (sums[i])
        {
            printf("%d", i);
            break;
        }
    }

    return 0;
}