#include<bits/stdc++.h>
int n,a,sum;
std::bitset<4000010>f;
int main(){
	scanf("%d",&n);f[0]=1;
    for(;n;--n){scanf("%d",&a);sum+=a;f|=f<<a;}
	for(sum=(sum+1)>>1;!f[sum];++sum);
	printf("%d",sum);
}
### 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>
int n,a,sum;
std::bitset<4000010>f;
int main(){
	scanf("%d",&n);f[0]=1;
    for(;n;--n){scanf("%d",&a);sum+=a;f|=f<<a;}
	for(sum=(sum+1)>>1;!f[sum];++sum);
	printf("%d",sum);
}
```