#include <bits/stdc++.h>

using namespace std;

bitset<4000010>b;
int N,A[2010];

int main()
{
	b.reset();
	b.set(0);
	scanf("%d",&N);
	int S = 0;
	for (int i=1;i<=N;i++)scanf("%d",&A[i]),b |= (b << A[i]),S += A[i];
	for (int i=(S + 1) / 2; i <= S; i++)
		if (b[i]){
			printf("%d\n",i);
			break;
		}
}
### 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<4000010>b;
int N,A[2010];

int main()
{
	b.reset();
	b.set(0);
	scanf("%d",&N);
	int S = 0;
	for (int i=1;i<=N;i++)scanf("%d",&A[i]),b |= (b << A[i]),S += A[i];
	for (int i=(S + 1) / 2; i <= S; i++)
		if (b[i]){
			printf("%d\n",i);
			break;
		}
}
```