#include <bitset>
#include <iostream>
using namespace std;

int main() {
    int n; cin >> n;
    bitset<4000000> b = 1;
    int sum = 0;
    while (n--) { int a; cin >> a; b = b | (b << a); sum += a; }
    for (int r = (sum+1)/2; ; r++) if (b.test(r)) { cout << r << endl; break; }
}

### Prompt

Please formulate a cpp solution 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 <iostream>
using namespace std;

int main() {
    int n; cin >> n;
    bitset<4000000> b = 1;
    int sum = 0;
    while (n--) { int a; cin >> a; b = b | (b << a); sum += a; }
    for (int r = (sum+1)/2; ; r++) if (b.test(r)) { cout << r << endl; break; }
}

```