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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#define _USE_MATH_DEFINES
#pragma region include
#include <iostream>
#include <iomanip>
#include <stdio.h>

#include <sstream>
#include <algorithm>
#include <iterator>
#include <cmath>
#include <complex>

#include <string>
#include <cstring>
#include <vector>
#include <bitset>

#include <queue>
#include <set>
#include <map>
#include <stack>
#include <list>

#include <ctime>
////
#include <random>//
#pragma endregion //#include
/////////

#pragma region typedef
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
#pragma endregion //typedef
////定数
const int INF = (int)1e9;
const LL MOD = (LL)1e9+7;
const LL LINF = (LL)4e18+20;
const LD PI = acos(-1.0);
const double EPS = 1e-9;
/////////
using namespace::std;

void solve(){
	int N;
	cin >> N;
	vector<int> A(N);
	int sum = 0;
	for(int i=0;i<N;++i){
		cin>>A[i];
		sum += A[i];
	}
	sort(A.begin(),A.end());
	int MAX = 0;
	vector<bool> dp(2000*2000+2010,false);
	dp[0] = true;
	for(int i=0;i<N;++i){
		for(int j=MAX;j>=0;--j){
			dp[j+A[i]] = dp[j+A[i]] || dp[j];
			MAX = max(MAX,(j+A[i])*dp[j+A[i]]);
		}
	}
	for(int i=(sum+1)/2;;++i){
		if( dp[i] ){
			cout << i << endl;
			return;
		}
	}
}
#pragma region main
signed main(void){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	std::cout << std::fixed;//小数を10進数表示
	cout << setprecision(16);//小数点以下の桁数を指定//coutとcerrで別	

	solve();
}
#pragma endregion //main()