#include<iostream>
#include<bitset>
using namespace std;
int main(){
	int n,m;
	int i,j,k;
	int a,b,c;
	bitset<4000005> bs(1);
	cin>>n;
	for(i=0,b=0;i<n;i++){
		cin>>a;
		b+=a;
		bs|=(bs<<a);
	}
	for(i=(b+1)/2;i<b;i++){
		if(bs[i]==1){a=i;break;}
	}
	cout<<a<<endl;
	return 0;
}
	

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

Examples

Input

3
1 2 1


Output

2


Input

1
58


Output

58

### Response

```cpp
#include<iostream>
#include<bitset>
using namespace std;
int main(){
	int n,m;
	int i,j,k;
	int a,b,c;
	bitset<4000005> bs(1);
	cin>>n;
	for(i=0,b=0;i<n;i++){
		cin>>a;
		b+=a;
		bs|=(bs<<a);
	}
	for(i=(b+1)/2;i<b;i++){
		if(bs[i]==1){a=i;break;}
	}
	cout<<a<<endl;
	return 0;
}
	

```