#include <bits/stdc++.h>
using namespace std;
bitset<4000005> dp;
int a[2005], n, s;
int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) scanf("%d", a + i), s += a[i];
	dp[0] = 1;
	for(int i = 1; i <= n; i++) dp |= dp << a[i];
	for(int i = (s + 1) / 2; i <= s; i++) {
		if(dp[i]) return printf("%d\n", i), 0;
	}
}
### Prompt

Generate 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;
bitset<4000005> dp;
int a[2005], n, s;
int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) scanf("%d", a + i), s += a[i];
	dp[0] = 1;
	for(int i = 1; i <= n; i++) dp |= dp << a[i];
	for(int i = (s + 1) / 2; i <= s; i++) {
		if(dp[i]) return printf("%d\n", i), 0;
	}
}
```