#include<cstdio>
#include<algorithm>
#include<bitset>
using namespace std;
int main(){
	bitset<5000005> dp;
	int n;
	scanf("%d",&n);
	int sum=0;
	dp[0]=1;
	for (int i=1; i<=n; i++) {
		int x;
		scanf("%d",&x);
		dp|=dp<<x;
		sum+=x;
	}
	for (int i=(sum+1)/2; i; i++)
		if (dp[i]) {
			printf("%d",i);
			return 0;
		}
}
### Prompt

Create a solution in CPP for 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<cstdio>
#include<algorithm>
#include<bitset>
using namespace std;
int main(){
	bitset<5000005> dp;
	int n;
	scanf("%d",&n);
	int sum=0;
	dp[0]=1;
	for (int i=1; i<=n; i++) {
		int x;
		scanf("%d",&x);
		dp|=dp<<x;
		sum+=x;
	}
	for (int i=(sum+1)/2; i; i++)
		if (dp[i]) {
			printf("%d",i);
			return 0;
		}
}
```