//http://judge.openfmi.net:9280/practice/get_problem_description?contest_id=2&problem_id=4
#include <stdio.h>
#include <algorithm>

int main()
{
	int n;
	scanf("%d", &n);
	
	int *arr = new int [n];
	int i, j, sum = 0;
	for (i = 0; i < n; ++i)
	{
		scanf("%d", arr + i);
		sum += arr[i];
	}
	int target = sum / 2;
	
	// curr[s] = can we make a sum = s with current number of elements[0..j]
	// prev[s] = can we make a sum = s with current number of elements[0..j - 1]
	bool *prev = new bool[target + 1]();
	bool *curr = new bool[target + 1];
	
	for (j = 1; j <= n; ++j)
	{
		prev[0] = true;
		for (i = 1; i <= target; ++i)
		{
			// we can make the sum if we could previously
			curr[i] = prev[i];
			if (arr[j - 1] <= i)
				curr[i] ||= prev[i - arr[j - 1]];
				// a[0] + a[1] + .. + a[j - 1] == sum <=> a[0] + a[1] + .. + a[j - 2] == sum - a[j - 1]
				// sum = i
				// we can make sum = i with elements [1..j-1], if we could make
				// sum = i - arr[j-1] with elements [1..j-2] (and adding arr[j-1])
				
		}
		std::swap(prev, curr);
	}
	
	for (i = target; i >= 0; --i)
	{
		if (prev[i])
		{
			printf("%d\n", (sum - i) - i);
			break;
		}
	}
}