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)

#include<bits/stdc++.h>
#include<bitset>

const int N = 2000 * 2000 + 1;
using value_type = long long;

std::bitset<N> bit;
value_type sum = 0;
int n;

inline int read()
{
	int s = 0,f = 1; char ch;
	while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
	while(isdigit(ch)) s = s * 10 + ch - '0',ch = getchar();
	return s * f;
}

int main()
{
	bit[0] = 1;
	n = read();
	for(int i = 1,x;i <= n;i++)
		x = read(),bit |= (bit << x),sum += x;
	for(auto i = (sum + 1) >> 1;i <= sum;i++)
		if(bit[i]) {
			printf("%lld",i);
			return 0;
		}

	exit(0);
}