#include<bits/stdc++.h>
using namespace std;

bitset<4000020> b;

int n,a,c=0,cc;

int main()
{
	scanf("%d",&n);
	b[0]=1;
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&a);
		b|=b<<a,c+=a;
	}
	cc=c;
	c=(c+1)>>1;
	for(int i=c;i<=cc;++i)
	{
		if(b[i])
		{
			printf("%d\n",i);
			return 0;
		}
	}
	
	return 0;
}
### Prompt

Generate a CPP solution to 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<bits/stdc++.h>
using namespace std;

bitset<4000020> b;

int n,a,c=0,cc;

int main()
{
	scanf("%d",&n);
	b[0]=1;
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&a);
		b|=b<<a,c+=a;
	}
	cc=c;
	c=(c+1)>>1;
	for(int i=c;i<=cc;++i)
	{
		if(b[i])
		{
			printf("%d\n",i);
			return 0;
		}
	}
	
	return 0;
}
```