#include <bits/stdc++.h>
using namespace std;
const int N=2002;
const int B=4000000+1;
int n,a[N],sum=0;
bitset<B> f;
int main(){
	scanf("%d",&n);
	f[0]=1;
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		f=f|(f<<a[i]);
		sum+=a[i];
	}
	for(int i=(sum+1)/2;;i++)if(f[i]){
		return !printf("%d\n", i);
	}

}
/*
1 3
0 1 3 4
*/
### Prompt

Develop a solution in cpp to the problem described below:

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=2002;
const int B=4000000+1;
int n,a[N],sum=0;
bitset<B> f;
int main(){
	scanf("%d",&n);
	f[0]=1;
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		f=f|(f<<a[i]);
		sum+=a[i];
	}
	for(int i=(sum+1)/2;;i++)if(f[i]){
		return !printf("%d\n", i);
	}

}
/*
1 3
0 1 3 4
*/
```