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)

#include<bits/stdc++.h>

using namespace std;

#define rep(i,a,b) for(int i = a; i<int(b);++i)
#define all(v) v.begin(),v.end()
#define sz(v) v.size()
#define trav(a,c) for(auto a: c)

typedef long long ll;
typedef vector<ll> vi;
typedef pair<ll,ll> pii;

int main(){
	cin.sync_with_stdio(false);
    vi v(2001);
    ll n; cin>>n;
    ll mxA = 0;
    rep(i,0,n){
        ll a;
        cin>>a;
        mxA = max(mxA,a);
        v[a]++;
    }

    if(v[mxA]%2==1){
        cout<<mxA*(v[mxA]/2+1)<<endl;
        return 0;
    }

    ll sum = 0;

    for(ll i = 2000;i>=0;--i){
        if(v[i]%2==1){
            cout<<sum + i*(v[i]/2+1)<<endl;
            return 0;
        }
        sum += (v[i]/2)*i;
    }
    cout<<sum<<endl;
}