#include<cstdio>
#include<bitset>
std::bitset<4000032> f;
int main()
{
	int N,sum=0,a,i,k;
	f[0] = 1;
	scanf("%d",&N);
	for(i=0;i<N;i++)scanf("%d",&a),f |= (f<<a),sum += a;
	k = (sum+1)>>1;
	while(f[k] == 0)k++;
	printf("%d",k);
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
std::bitset<4000032> f;
int main()
{
	int N,sum=0,a,i,k;
	f[0] = 1;
	scanf("%d",&N);
	for(i=0;i<N;i++)scanf("%d",&a),f |= (f<<a),sum += a;
	k = (sum+1)>>1;
	while(f[k] == 0)k++;
	printf("%d",k);
	return 0;
}
```