#include<bits/stdc++.h>
using namespace std;
const int N=2e3+5;
bitset<N*N>b;
int n,a[N],i,s,ss;
int main(){
	b[0]=1;
	scanf("%d",&n);for(i=1;i<=n;++i)scanf("%d",a+i),b|=b<<a[i];
	for(i=1;i<N*N;++i)s+=b[i];
	for(i=1;ss+=b[i],ss<s+1>>1;++i);printf("%d\n",i);
}
### 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<bits/stdc++.h>
using namespace std;
const int N=2e3+5;
bitset<N*N>b;
int n,a[N],i,s,ss;
int main(){
	b[0]=1;
	scanf("%d",&n);for(i=1;i<=n;++i)scanf("%d",a+i),b|=b<<a[i];
	for(i=1;i<N*N;++i)s+=b[i];
	for(i=1;ss+=b[i],ss<s+1>>1;++i);printf("%d\n",i);
}
```