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.util.*;
import java.io.*;
import java.math.BigInteger;
//267630EY
public class Main
{
  static PrintWriter out=new PrintWriter(System.out);

  public static void main(String[] args) throws IOException
  {
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    int[] a=sc.nextIntArray(n);
    int sum=0;
    int ans=0;
    for(int i=0;i<n;i++)
    {
        sum+=a[i];
    }
    BitSet possible=new BitSet(4000001);
    possible.set(4000000);
    for(int x:a)
    {
        possible.or(possible.get(x,4000001));
    }
    ans=possible.previousSetBit(4000000-(sum+1)/2);

    out.println(4000000-ans);
    

    out.flush();



        
   
      
      
                
    
  }
  
  static class Scanner
  {
    BufferedReader br;
    StringTokenizer tk=new StringTokenizer("");
    public Scanner(InputStream is) 
    {
      br=new BufferedReader(new InputStreamReader(is));
    }
    
    public boolean hasNext()
    {
      

      boolean result=false;
      
      try 
      {
        result = br.ready();
      } 
      catch (IOException e)
      {
        System.err.println(e);
      }
      return result;
    }
    
    public int nextInt() throws IOException
    {
      if(tk.hasMoreTokens())
        return Integer.parseInt(tk.nextToken());
      tk=new StringTokenizer(br.readLine());
      return nextInt();
    }
    public long nextLong() throws IOException
    {
      if(tk.hasMoreTokens())
        return Long.parseLong(tk.nextToken());
      tk=new StringTokenizer(br.readLine());
      return nextLong();
    }
    public String next() throws IOException
    {
      if(tk.hasMoreTokens())
        return (tk.nextToken());
      tk=new StringTokenizer(br.readLine());
      return next();
    }
    public String nextLine() throws IOException
    {
      tk=new StringTokenizer("");
      return br.readLine();
    }
    public double nextDouble() throws IOException
    {
      if(tk.hasMoreTokens())
        return Double.parseDouble(tk.nextToken());
      tk=new StringTokenizer(br.readLine());
      return nextDouble();
    }
    public char nextChar() throws IOException
    {
      if(tk.hasMoreTokens())
        return (tk.nextToken().charAt(0));
      tk=new StringTokenizer(br.readLine());
      return nextChar();
    }
    public int[] nextIntArray(int n) throws IOException
    {
      int a[]=new int[n];
      for(int i=0;i<n;i++)
        a[i]=nextInt();
      return a;
    }
    public long[] nextLongArray(int n) throws IOException
    {
      long a[]=new long[n];
      for(int i=0;i<n;i++)
        a[i]=nextLong();
      return a;
    }
    public int[] nextIntArrayOneBased(int n) throws IOException
    {
      int a[]=new int[n+1];
      for(int i=1;i<=n;i++)
        a[i]=nextInt();
      return a;
    }
    public long[] nextLongArrayOneBased(int n) throws IOException
    {
      long a[]=new long[n+1];
      for(int i=1;i<=n;i++)
        a[i]=nextLong();
      return a;
    }
  }
}