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)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define pb push_back
#define fs first
#define sc second
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<double> vd;
typedef pair<int, int> pii;
typedef pair<ll, int> plli;
typedef vector<pii> vpii;
typedef tree<pii, null_type, less<pii>, rb_tree_tag,
    tree_order_statistics_node_update> ost;

const int N=2000;
const int R=2000001;
int a[N];
bitset<R> x, y;

int main() {
    int i, s, n;
    scanf("%d", &n);
    for(s=i=0; i<n; i++) {
        scanf("%d", &a[i]);
        s+=a[i];
    }
    x|=1;
    for(i=0; i<n; i++) {
        y=x<<a[i];
        x|=y;
    }
    for(i=s/2; !x.test(i); i--);
    printf("%d\n", s-i);
    return 0;
}