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 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define ll long long
#define re register
#define inf 0x3f3f3f3f
using namespace std;
bitset<4000007>f;
inline int read(){
   int x=0,w=0;char ch=getchar();
   while (!isdigit(ch))w|=ch=='-',ch=getchar();
   while (isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
   return w?-x:x;
}
int main(){
	int sum=0;
	int n=read();f[0]=1;
	for (int i=1;i<=n;++i){
		int x=read();
		f|=f<<x;
		sum+=x;
	}
	for (int i=sum/2;i<=sum;++i)
		if (f[i]){
			printf("%d",i);
			return 0;
		}
	return 0;
}