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)

// In the name of God

#include <iostream>
#include <algorithm>
#include <fstream>
#include <vector>
#include <deque>
#include <assert.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <stdio.h>
#include <string.h>
#include <utility>
#include <math.h>
#include <bitset>
#include <iomanip>
#include <complex>

using namespace std;

#define rep(i, a, b) for (int i = (a), i##_end_ = (b); i < i##_end_; ++i)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define x first
#define y second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool smin(T &a, const T &b)   { return a > b ? a = b : a;    }
template<typename T> inline bool smax(T &a, const T &b)   { return a < b ? a = b : a;    }

typedef long long LL;

const int N = (int) 2005 * 2005, mod = (int) 0;
bitset<N> dp;
int main() {
	dp[0] = 1;
	int n, sum = 0;;
	cin >> n;
	for (int j = 0; j < n; ++j) {
		int x;
		cin >> x;
		sum += x;
		dp |= dp << x;
	}
	for (int j = 0; j < N; ++j)
		if (dp[j] && j + j >= sum) {
			cout << j << endl;
			return 0;
		}
}