#include <bits/stdc++.h>

using namespace std;

int n, sum;
bitset<4000010> dp;

int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	dp.set(0);
	for (int i = 1; i <= n; i++)
	{
		int x; cin >> x;
		sum += x;
		dp |= (dp << x);
	}
	for (int i = (sum + 1) / 2; ; i++)
		if (dp[i])
		{
			cout << i << endl;
			break;
		}
	return 0;
}

### Prompt

Your challenge is to write 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 <bits/stdc++.h>

using namespace std;

int n, sum;
bitset<4000010> dp;

int main()
{
	ios::sync_with_stdio(false);
	cin >> n;
	dp.set(0);
	for (int i = 1; i <= n; i++)
	{
		int x; cin >> x;
		sum += x;
		dp |= (dp << x);
	}
	for (int i = (sum + 1) / 2; ; i++)
		if (dp[i])
		{
			cout << i << endl;
			break;
		}
	return 0;
}

```