Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
1 2 1

Sample Output 1

2

In this case, S = (1, 1, 2, 2, 3, 3, 4). Its median is S_4 = 2.

Sample Input 2

1
58

Sample Output 2

58

In this case, S = (58).

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef unsigned long long u64;

#define N (62500)

u64 dp[N + 1000];

int main(void) {
  int n;
  scanf("%d", &n);
  
  dp[0] = 1;

  int s = 0;
  while (n--) {
    int a;
    scanf("%d", &a);

    s += a;

    int H = a / 64;
    int L = a % 64;

    if (L == 0) {
      for (int i = N; i >= 0; i--) {
        dp[i + H] |= dp[i];
      }
    } else {
      for (int i = N; i >= 0; i--) {
        dp[i + H + 1] |= dp[i] >> (64 - L);
        dp[i + H] |= dp[i] << L;
      }
    }
  }

  s = (s + 1) / 2;
  while (~dp[s / 64] >> s % 64 & 1) s++;

  printf("%d\n", s);

  return 0;
}