#include<bits/stdc++.h>
#define ll long long
#define N 100009
using namespace std;

bitset<4000009> a;
int n;
int main(){
	scanf("%d",&n);
	int i,x,sum=0;
	a[0]=1;
	for (i=1; i<=n; i++){
		scanf("%d",&x); sum+=x;
		a|=a<<x;
	}
	for (i=sum+1>>1; i<=sum; i++) if (a[i]) break;
	printf("%d\n",i);
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
#define ll long long
#define N 100009
using namespace std;

bitset<4000009> a;
int n;
int main(){
	scanf("%d",&n);
	int i,x,sum=0;
	a[0]=1;
	for (i=1; i<=n; i++){
		scanf("%d",&x); sum+=x;
		a|=a<<x;
	}
	for (i=sum+1>>1; i<=sum; i++) if (a[i]) break;
	printf("%d\n",i);
	return 0;
}

```