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

bitset<2000 * 2000 + 10> b;

int main() {
    int n, x, s = 0;
    cin >> n; b.set(0);
    while (n--) {
        cin >> x; b |= (b << x), s += x;
    }
    for (int i = (s + 1) / 2; ; i++) {
        if (b.test(i)) printf("%d\n", i), exit(0);
    }
    return 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;

bitset<2000 * 2000 + 10> b;

int main() {
    int n, x, s = 0;
    cin >> n; b.set(0);
    while (n--) {
        cin >> x; b |= (b << x), s += x;
    }
    for (int i = (s + 1) / 2; ; i++) {
        if (b.test(i)) printf("%d\n", i), exit(0);
    }
    return 0;
}
```