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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=2001,MAXsum=2000001,INF=1<<30;
vector<int> A(MAX),dp(MAXsum);
int N;

int check(int rem,int now){
    if(rem<0) return 0;
    if(rem==0) return 1;
    if(now==N) return 0;
    if(dp[rem]) return 1;
    return dp[rem]=(check(rem-A[now],now+1)||check(rem,now+1));
}

int main(){
    
    cin>>N;
    int sum=0,right;
    for(int i=0;i<N;i++){
        cin>>A[i];
        sum+=A[i];
    }
    right=sum/2;
    
    for(int i=right;i>=0;i--){
        if(check(i,0)){
            cout<<sum-i<<endl;
            break;
        }
    }
}