#include <bitset>
#include <cstdio>
int n, a, s; std::bitset<4000001> b = 1;
int main() {
	scanf("%d", &n); 
	for (int i = 0; i < n; i++) scanf("%d", &a), s += a, b |= b << a;
	for (int i = s; i >= (s + 1) / 2; i--) if (b[i]) a = i;
	printf("%d\n", a);
}
### Prompt

Please create a solution in CPP to 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 <bitset>
#include <cstdio>
int n, a, s; std::bitset<4000001> b = 1;
int main() {
	scanf("%d", &n); 
	for (int i = 0; i < n; i++) scanf("%d", &a), s += a, b |= b << a;
	for (int i = s; i >= (s + 1) / 2; i--) if (b[i]) a = i;
	printf("%d\n", a);
}
```