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++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
 
using namespace std;
typedef unsigned long long Int;
 
Int ok[63000];
Int n;
Int a;
Int sum;
int main(){
	cin >> n;
	ok[0] = 1;
	for(Int i = 0;i < n;i++){
		cin >> a;
		sum += a;
		for(Int i = 62600;i >= 0;i--){
			if(ok[i] == 0)continue;
			Int tmp = ok[i];
			Int shift = a % 64;
			Int plus = a / 64;
			ok[i + plus] |= tmp << shift;
			ok[i + plus + 1] |= tmp >> (64 - shift);
			
			if(i == 0)break;
		}
	}
		
	Int med = sum / 2 + sum % 2;
 
	for(int i = med;i <= 4000000;i++){
		if((ok[i/64]>>(i%64))%2){
			cout << i << endl;
			return 0;
		}
	}
}