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

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.BigInteger;
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.flush();out.close();
    }
        static class TaskE {
         public void solve(int testNumber, InputReader in, PrintWriter out) {
             int n=in.nextInt();
             int s=0;int a[]=new int[n];
             for(int i=0;i<n;i++)s+=a[i]=in.nextInt();
             BigInteger bi=BigInteger.ONE;
             for(int i=0;i<n;i++){
              bi=bi.or(bi.shiftLeft(a[i]));
             }
             if(s%2==0)s=s/2;else s=s/2+1;//System.out.println(s);
             for(int i=s;;i++){
              if(bi.testBit(i)){out.print(i);return;}
             }
         }
    }
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;
        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return null;
                st = new StringTokenizer(s);
            }
            return st.nextToken();
        }

        boolean hasMoreTokens() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return false;
                st = new StringTokenizer(s);
            }
            return true;
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}