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: Wrong Answer)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;

template <typename T1, typename T2>
string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
    bool done_something = false;
    stringstream res;
    res << "[";
    for (; begin_iter != end_iter and counter; ++begin_iter) {
        done_something = true;
        counter--;
        res << *begin_iter << ", ";
    }
    string str = res.str();
    if (done_something) {
        str.pop_back();
        str.pop_back();
    }
    str += "]";
    return str;
}

vector<int> SortIndex(int size, std::function<bool(int, int)> compare) {
    vector<int> ord(size);
    for (int i = 0; i < size; i++) ord[i] = i;
    sort(ord.begin(), ord.end(), compare);
    return ord;
}

template <typename T>
bool MinPlace(T& a, const T& b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template <typename T>
bool MaxPlace(T& a, const T& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template <typename Container>
int SZ(const Container& S) { return S.size(); }

template <typename S, typename T>
ostream& operator <<(ostream& out, const pair<S, T>& p) {
    out << "{" << p.first << ", " << p.second << "}";
    return out;
}

template <typename T>
ostream& operator <<(ostream& out, const vector<T>& v) {
    out << "[";
    for (int i = 0; i < (int)v.size(); i++) {
        out << v[i];
        if (i != (int)v.size()-1) out << ", ";
    }
    out << "]";
    return out;
}

#if DEBUG && !ONLINE_JUDGE
    ifstream input_from_file("input.txt");
    #define cin input_from_file

    #define dbg_var(x) clog << #x  << ": " << x << endl;
    #define dbg_array(x, len) clog << #x << ": " << print_iterable(x, x+len, -1) << endl;
#else
    #define dbg_var(x)
    #define dbg_array(x, len)
#endif

///////////////////////////////////////////////////////////////////////////
//////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
///////////////////////////////////////////////////////////////////////////

const int MAXN = 40010; 

bool done[2][2*MAXN];
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); // Remove in problems with online queries!
    int N;
    cin >> N;
    vector<int> a(N);
    int S = 0;
    for (int i = 0; i < N; i++) {
        cin >> a[i];
        S += a[i];
    }
    
    int zero = MAXN;
    done[0][zero] = true;
    random_shuffle(a.begin(), a.end());
    bool it = 0;
    for (int x : a) {
        for (int i = 0; i < 2*MAXN; i++) done[it^1][i] = 0;
        for (int i = 0; i < 2*MAXN; i++) {
            if (i + x < 2*MAXN) done[it^1][i+x] |= done[it][i];
            if (i - x >= 0) done[it^1][i-x] |= done[it][i];
        }
        it ^= 1;
    }
    for (int i = 0; i < MAXN; i++) {
        if (done[it][zero+i]) {
            cout  << (S+i)/2 << "\n";
            break;
        }
    }
}