#include<bits/stdc++.h>
using namespace std;
const int maxn=2010;
bitset<2000010> f;
int a[maxn];
int n,sum;
int main(){
	cin >> n;
	for(int i=1;i<=n;i++)
		cin >> a[i],sum+=a[i];
	f[0]=1;
	for(int i=n;i>=1;--i)
		f|=f<<a[i];
	for(int i=(sum+1)>>1;i<=sum;++i)
		if(f[i]){
			cout << i << endl;
			return 0;
        }
    return 0;
}
### 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<bits/stdc++.h>
using namespace std;
const int maxn=2010;
bitset<2000010> f;
int a[maxn];
int n,sum;
int main(){
	cin >> n;
	for(int i=1;i<=n;i++)
		cin >> a[i],sum+=a[i];
	f[0]=1;
	for(int i=n;i>=1;--i)
		f|=f<<a[i];
	for(int i=(sum+1)>>1;i<=sum;++i)
		if(f[i]){
			cout << i << endl;
			return 0;
        }
    return 0;
}
```