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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;
using System.Text;

class Solution
{
    static void Main(string[] args)
    {
        int n = int.Parse(Console.ReadLine());
        int[] a = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
        int s = a.Sum();
        BitArray[] ba = new BitArray[2];
        ba[0] = new BitArray(2000 * 2000 + 1, false);
        ba[1] = new BitArray(2000 * 2000 + 1, false);
        ba[0][0] = true;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j <= 2000 * 2000; j++)
            {
                ba[~i & 1][j] |= ba[i & 1][j];
                if (j + a[i] <= 2000 * 2000)
                    ba[~i & 1][j + a[i]] |= ba[i & 1][j];
            }
        }

        for (int i = 0; i <= 2000 * 2000; i++)
        {
            if (i * 2 < s) continue;
            if (ba[n & 1][i])
            {
                Console.WriteLine(i);
                return;
            }
        }
    }
}