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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define INF 1000000000
#define LINF 1000000000000000000
#define MOD 1000000007
#define mod 998244353
#define INF63 1061109567
#define INF127 9187201950435737471
#define UINF 18446744073709551615
#define F first
#define S second
#define ll long long
#define N 2010
using namespace std;
bitset<N*N> dp;
ll n,a[N],sum=0; 
int main(){
	ll i;
	scanf("%lld",&n);
	for(i=0;i<n;i++)
	{
		scanf("%lld",&a[i]);
		sum+=a[i];
	}
	sum=(sum+1)/2;
	dp[0]=1;
	for(i=0;i<n;i++)
	{
		dp|=(dp<<a[i]);
	}
	for(i=sum;i<=sum*2;i++)
	{
		if(dp[i]==1)
		{
			printf("%lld\n",i);
			return 0;
		}
	}
	return 0;
}