#include <bits/stdc++.h>
using namespace std;

int n, a, J;
bitset<4000006> b;

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	b[0] = 1;
	for(int i = 1; i <= n; ++i) {
		cin >> a;
		b |= b << a;
		J += a;
	}
	for(int i = (J + 1) / 2; true; ++i) {
		if(b[i]) {
			cout << i << endl;
			return 0;
		}
	}
	return 0;
}
### Prompt

In Cpp, your task is to solve 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 <bits/stdc++.h>
using namespace std;

int n, a, J;
bitset<4000006> b;

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n;
	b[0] = 1;
	for(int i = 1; i <= n; ++i) {
		cin >> a;
		b |= b << a;
		J += a;
	}
	for(int i = (J + 1) / 2; true; ++i) {
		if(b[i]) {
			cout << i << endl;
			return 0;
		}
	}
	return 0;
}
```