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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define mkp make_pair
#define pb push_back
#define loop(i,n)   for(ll i=0;i<n;i++)
#define rloop(i,n)  for(ll i=n-1;i>=0;i--)
#define pii pair<ll,ll>
#define pll pair<ll,ll>
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);  cout.tie(NULL);
#define test ll t; cin>>t; while(t--)
#define F first
#define S second
#define all(a) (a.begin(),a.end())
#define pi 3.1415926

int main()
{
	fast
	//test
	{
		ll n,sum=0;	cin>>n;
		vll arr(n);
		loop(i,n)
		{
			cin>>arr[i];
			sum+=arr[i];
		}
		vector<vll> dp(n+1,vll (sum+1));
		loop(i,n+1)
		{
			loop(j,sum+1)
			{
				if(j==0)
					dp[i][j]=1;
				else if(i==0)
					dp[i][j]=0;
				else
				{
					dp[i][j]=dp[i-1][j];
					if(arr[i-1]<=j)
						dp[i][j]=dp[i-1][j-arr[i-1]];
				}
			}
		}
		for(int i=sum/2;i>=0;i--)
		{
			if(dp[n][i])
			{
				cout<<sum-i<<endl;
				break;
			}
		}
	}
}