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>
using namespace std;
const int N=2050;
inline int gi(){
  char ch=getchar();int x=0,q=0;
  while(ch<'0' || ch>'9') ch=='-'?q=1:0,ch=getchar();
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return q?(-x):x;
}
int a[N],ans;
bitset<4000001>f;
int main(){
  int n=gi(),mn=1e9,p;
  f.set(0);
  for (int i=1; i<=n; ++i){
    a[i]=gi(),ans+=a[i];
    f|=f<<a[i];
  }
  for (int i=0; i<=ans; ++i)
    if (f[i]==1)
      if (abs(ans-2*i)<mn)
	mn=abs(ans-2*i),p=max(i,ans-i);
  cout<<p;
  return 0;
}