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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using static System.Console;
namespace AtCoder.AGC020
{
    class MedianSum
    {
        public static void Main()
        {
            int n = int.Parse(ReadLine());
            var aa = ReadLine().Trim().Split(' ').Select(_ => int.Parse(_));
            var ss = new Queue<int>();
            var memo = new int[aa.Sum() + 1];
            memo[0] = 0;
            ss.Enqueue(0);
            foreach(var a in aa)
            {
                for (var i = ss.Count(); i > 0; --i)
                {
                    if (memo[a + ss.Peek()] == 0) ss.Enqueue(a + ss.Peek());
                    ++memo[a + ss.Peek()]; ss.Enqueue(ss.Peek()); ss.Dequeue();
                }
            }
            var left = 0; var right = memo.Length - 1;
            var tend_left = 0;
            while (left < right)
            {
                if (tend_left >= 0) tend_left -= memo[right--];
                else tend_left += memo[left++];
            }
            WriteLine((tend_left + memo[left] >= 0) ? left : left + 1);
        }
    }
}