#include<cstdio>
#include<bitset>
using namespace std;
const int N=4e6+7;
bitset<N>s;int n,i,x,sum;
int main(){
	s[0]=1;
	for(scanf("%d",&n),i=1;i<=n;++i)scanf("%d",&x),s|=(s<<x),sum+=x;
	for(x=(sum+1)/2;;++x)if(s[x])break;
	printf("%d\n",x);
}
### Prompt

Create a solution in Cpp for the following problem:

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<cstdio>
#include<bitset>
using namespace std;
const int N=4e6+7;
bitset<N>s;int n,i,x,sum;
int main(){
	s[0]=1;
	for(scanf("%d",&n),i=1;i<=n;++i)scanf("%d",&x),s|=(s<<x),sum+=x;
	for(x=(sum+1)/2;;++x)if(s[x])break;
	printf("%d\n",x);
}
```