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: Accepted)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author tatsumack
 */

#include <iostream>
#include <fstream>

#include <iostream>
#include <limits.h>
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>

#include <list>
#include <map>
#include <numeric>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <unordered_map>
#include <unordered_set>

#define int long long
#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define REPS(i, n) for (int i = 1, i##_len = (n); i <= i##_len; ++i)
#define FOR(i, a, b) for (int i = (a), i##_len = (b); i <= i##_len; ++i)
#define REV(i, a, b) for (int i = (a); i >= (b); --i)
#define CLR(a, b) memset((a), (b), sizeof(a))
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define INF 1001001001001001001ll
#define fcout cout << fixed << setprecision(12)

using namespace std;

typedef pair<int, int> P;

class CMedianSum {
public:
    static constexpr int kStressIterations = 0;

    static void generateTest(std::ostream& test) {
    }

    void solve(std::istream& cin, std::ostream& cout) {
        int N, sum = 0;
        cin >> N;
        vector<int> A(N);
        REP(i, N) {
            cin >> A[i];
            sum += A[i];
        }

        const int MAX = 2005 * 2005;
        bitset<MAX> b;
        b[0] = 1;
        REP(i, N) {
            b |= b << A[i];
        }

        FOR(i, (sum + 1) / 2, sum) {
            if (b[i] == 1) {
                cout << i << endl;
                return;
            }
        }

    }
};


signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    std::istream& in(std::cin);
    std::ostream& out(std::cout);

    CMedianSum solver;
    solver.solve(in, out);
    return 0;
}