#include<bits/stdc++.h>

using namespace std;

const int N=2005;

bitset<N*N>f;

int main(){
    int n,s=0;scanf("%d",&n);
    f[0]=1;
    for(int i=1,x;i<=n;++i)scanf("%d",&x),f|=f<<x,s+=x;
    for(int i=(s+1)/2;;++i)if(f[i])return printf("%d\n",i),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<bits/stdc++.h>

using namespace std;

const int N=2005;

bitset<N*N>f;

int main(){
    int n,s=0;scanf("%d",&n);
    f[0]=1;
    for(int i=1,x;i<=n;++i)scanf("%d",&x),f|=f<<x,s+=x;
    for(int i=(s+1)/2;;++i)if(f[i])return printf("%d\n",i),0;
}


```