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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
// type
typedef long long ll;
typedef long double ld;
// loop
//#define For(i, l, r, x)     for (int i = l; i < r; i+=x)
//#define ForE(i, l, r, x)    for (int i = l; i <= r; i+=x)
//#define Ford(i, r, l)       for (int i = r; i > l; i--)
//#define FordE(i, r, l)      for (int i = r; i >= l; i--)
//#define Fora(i, a)          for (auto i : a)
// I/O 
#define FAST_IO             std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
//#define PrintV(a)           Fora(ii, a) cout << ii << ' '; cout << rl;
//#define PrintVl(a)          Fora(ii, a) cout << ii << rl;
//#define PrintA(a, l, r)     for (int ii = l; ii <= r; ii++) cout << a[ii] << ' '; cout << rl;
//#define PrintAl(a, l, r)    for (int ii = l; ii <= r; ii++) cout << a[ii] << rl;
//#define Ptest(x)            return cout << x, 0;
#define setpre(n)           fixed << setprecision(n)
// pair
#define F                   first
#define S                   second
#define pii                 pair<int, int>
#define pll                 pair<ll, ll>
#define pdd                 pair<ld, ld>
// vector & !!?(string)
#define eb                  emplace_back
#define pb                  push_back
#define all(a)              a.begin(), a.end()
#define rall(a)             a.rbegin(), a.rend()
#define sz(a)               a.size()
#define len(a)              a.length()
// geometry calc    
#define pi                  acos(-1.0)
#define g_sin(a)            sin(a*pi/180)
#define g_cos(a)            cos(a*pi/180)
#define g_tan(a)            tan(a*pi/180)
// set val
#define ms0(a)              memset(a,        0, sizeof(a));
#define ms1(a)              memset(a,        1, sizeof(a));
#define msn1(a)             memset(a,       -1, sizeof(a));
#define msinf(a)            memset(a, 0x3f3f3f, sizeof(a));
// constant
const int mod1 = 998244353, mod = 1e9+7;
const int MAXN = 100005, MAX_M = 200010;
// code
//#define int long long
const int X = (1 << 25);
bitset<X> dp;
void Solve() {
    int n, s = 0; cin >> n; dp[0] = 1;
    for (int i = 1, x; i <= n; i++){
      	cin >> x; s += x; dp = dp | (dp << x);
    }
    //cout << s << endl;
  	for (int ans = (s+1)/2; ans <= s; ans ++)
      	if (dp[ans]) {cout << ans; break;}
}

signed main(){
    FAST_IO;
    int TC = 1; 
	//cin >> TC;
    while(TC--) Solve();
    return 0;
}