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.Arrays;
import java.util.Scanner;

public class Main {
  static Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {
    int N = sc.nextInt();
    int[] A = new int[N];
    int sum = 0;
    for (int i = 0; i < N; i++) {
      A[i] = sc.nextInt();
      sum += A[i];
    }
    if (N == 1) {
      System.out.println(sum);
      return;
    }
    long[] exist = new long[(sum + 63) / 64];
    exist[0] = 1L;
    long[] tmp = new long[exist.length];
    for (int i = 0; i < N; i++) {
      Arrays.fill(tmp, 0L);
      for (int j = 0; j < exist.length; j++) {
        if (A[i] % 64 == 0) {
          if (j + A[i] / 64 < tmp.length) tmp[j + A[i] / 64] |= exist[j];
        } else {
          if (j + A[i] / 64 < tmp.length) tmp[j + A[i] / 64] |= exist[j] << (A[i] % 64);
          if (j + A[i] / 64 + 1 < tmp.length) tmp[j + A[i] / 64 + 1] |= exist[j] >>> (64 - A[i] % 64);
        }
        exist[j] |= tmp[j];
      }
      exist[A[i] / 64] |= 1L << (A[i] % 64);
    }

    for (int i = (sum + 1) / 2; ; i++) {
      if ((exist[i / 64] & (1L << i % 64)) != 0) {
        System.out.println(i);
        return;
      }
    }
  }

}