We will define the median of a sequence b of length M, as follows:

* Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the value of the (M / 2 + 1)-th element of b' is the median of b. Here, / is integer division, rounding down.



For example, the median of (10, 30, 20) is 20; the median of (10, 30, 20, 40) is 30; the median of (10, 10, 10, 20, 30) is 10.

Snuke comes up with the following problem.

You are given a sequence a of length N. For each pair (l, r) (1 \leq l \leq r \leq N), let m_{l, r} be the median of the contiguous subsequence (a_l, a_{l + 1}, ..., a_r) of a. We will list m_{l, r} for all pairs (l, r) to create a new sequence m. Find the median of m.

Constraints

* 1 \leq N \leq 10^5
* a_i is an integer.
* 1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the median of m.

Examples

Input

3
10 30 20


Output

30


Input

1
10


Output

10


Input

10
5 9 5 9 8 9 3 5 4 3


Output

8
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using Graph=vector<vector<ll>>;
#define MAX 200003
#define MOD 1000000007

template<class T>
class BIT{
  ll N;
  std::vector<T> bit;
public:
  BIT(ll n){
    N=n;
    bit.resize(N+1,0);
  }

  //i番目の要素にwを追加
  void add(ll i,T w){
    for(ll x=i;x<=N;x+=x&-x){
      bit[x]+=w;
    }
  }

  //[1,i]の和を求める
  T sum(ll i){
    ll ret=0;
    for(ll x=i;x>0;x-=x&-x){
      ret+=bit[x];
    }
    return ret;
  }
};


int main(){
  ll N;
  cin>>N;
  vector<ll> a(N);
  for(ll i=0;i<N;i++){
    cin>>a.at(i);
  }


  ll ok=0;
  ll ng=1000000001;
  while(ok+1<ng){
    ll x=(ok+ng)/2;
    vector<ll> sum(N,0);
    if(a.at(0)>=x){
      sum.at(0)++;
    }else{
      sum.at(0)--;
    }
    for(ll i=1;i<N;i++){
      if(a.at(i)>=x){
        sum.at(i)=sum.at(i-1)+1;
      }else{
        sum.at(i)=sum.at(i-1)-1;
      }
    }

    ll count=0;
    BIT<ll> tree(3*N);
    for(ll i=0;i<N;i++){
      count+=tree.sum(sum.at(i)+N+1);
      tree.add(sum.at(i)+N+1,1);
      if(sum.at(i)>=0){
        count++;
      }
    }

    if(count>=(N*(N+1)/2-1)/2+1){
      ok=x;
    }else{
      ng=x;
    }
  }

  cout<<ok<<endl;
}
