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)

///https://agc020.contest.atcoder.jp/tasks/agc020_c?lang=en

#include<bits/stdc++.h>
#define i64 long long
#define inf 1000000000000000000
using namespace std;

int ara[2001];
bitset<2001*2001>bset;//[2];
int main()
{
    int n;
    scanf("%d",&n);

    int sum=0,mx=0;
    for(int i=1; i<=n; i++)
        scanf("%d",&ara[i]);


    sort(ara+1,ara+n+1);

    for(int i=1; i<=n; i++)
    {
        sum+=ara[i];
        mx=max(mx,ara[i]);
    }

    int avg=(sum+2-1)/2, last=mx*n;

    bset[0]=1;
    for(int i=1; i<=n; i++)
        bset= bset | (bset<<ara[i]);


    for(int i=avg; i<=last; i++)
    {
        if(bset[i]==1)
        {
            printf("%d\n",i);
            return 0;
        }
    }


    return 0;
}