#include<bits/stdc++.h>
using namespace std;
const int N=4000010;
bitset<N>b;
int n,x,s;
int main()
{
	scanf("%d",&n);
    b[0]=1;
	for (int i=1;i<=n;i++) {scanf("%d",&x); b=((b<<x)|b);s+=x;}	
	if (s&1) s=s/2+1; else s/=2; 
	for (int i=s;i<N;i++) if (b[i]) {
		printf("%d\n",i); break;
	}
}
### Prompt

Please formulate 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;
const int N=4000010;
bitset<N>b;
int n,x,s;
int main()
{
	scanf("%d",&n);
    b[0]=1;
	for (int i=1;i<=n;i++) {scanf("%d",&x); b=((b<<x)|b);s+=x;}	
	if (s&1) s=s/2+1; else s/=2; 
	for (int i=s;i<N;i++) if (b[i]) {
		printf("%d\n",i); break;
	}
}
```