#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
#define ll long long
#define ld long double
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repo(i,n) for(int i = 1; i < (int)(n); i++)
#define pb push_back
#define mp make_pair
#define np next_permutation
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define uniq(v) v.erase(unique(v.begin(),v.end()),v.end())
#define lb(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define ub(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
const ll mod=1000000007;
//const ll mod=998244353;
const ld pi=acos(-1.0);
const ll INF = 1LL<<61;
template<class T>bool chmax(T &a, const T &b) { 
  if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) {
  if (b<a) { a=b; return 1; } return 0; }
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
ll jou(ll N, ll P){if(P==0) return 1;
  if(P%2==0){ll t = jou(N, P/2);return t*t % mod;
  }return (N * jou(N, P-1)) % mod;}  //jou(n,mod-2)で逆元
//intの最大値2147483647 ≒ 2×10^9
//long longの最大値9223372036854775807 ≒ 9×10^18
//'大文字'+=32;    で小文字に
//  cout << fixed << setprecision (20);   小数点以下2０桁まで
//実行時間制約2秒では２×10^8回くらいまで計算できる



//————————Fenwick  BIT  転倒数———————————————————————

template<typename T>
class BIT{
  public:
    int N;
    vector<T> data;
    BIT(T _N):N(_N){
        data.assign(N+1, 0);
    };
    
    // a is 1-indexed
    void add(int a, T w){
        for(int x = a; x <= N; x += x & -x) data[x] += w;
    }
    // 1-indexed sum of prefix [0, a]
    T sum(int a){
        T res = 0;
        for(int x = a; x > 0; x -= x & -x) res += data[x];
        return res;
    }
    // 1-indexed sum of range [l, r]
    T sum(int l, int r){return sum(r) - sum(l-1);}

    // 0-indexed add
    void add0(int a, T w){add(a + 1, w);}
    // 0-indexed sum
    T sum0(int a){return sum(a + 1);}
    // 0-indexed sum of range
    T sum0(int l, int r){return sum0(r) - sum0(l-1);}

    //1-indexed でv1+v2+...+vx >= w となる最小のx
    T blb(T w) {
      if(w <= 0) return 0;
      int res = 0;
      int i=1;
      while(i*2 <= N) {i *= 2;}
      //i は N以下の最大の２べき
      for(int k = i; k > 0; k /= 2) {
        if(res + k <= N && data[res + k] < w) {
          w -= data[res + k];
          res += k;
        }
      }
      return res+1;
    }

    /* debug */
    void print() {
        rep(i,data.size()) cout << sum0(i,i) << ",";
        cout << endl;
    }

};


  int tento(vector<ll>v){
    int n=v.size();

    //v2に座標圧縮
    vector<ll> v2(n);
    vector<pair<ll, int>> vp;    
    rep(i,n){
      vp.pb(mp(v[i], i));
    }
    sort(all(vp));
    rep(i,n) v2[vp[i].se]=i;

    BIT<ll> bit(n);
    int rtn = 0;
    rep(i,n){
      //rtn += bit.sum0(v2[i]) でi<jかつVi<Vjの組みの数が出せる
      rtn +=  i - bit.sum0(v2[i]);
      bit.add0(v2[i], 1);      
    }
    return rtn;
  } 


//——————————————————————————————————————————


int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin>>n;

  vector<ll> p(n);
  rep(i,n){
    cin>>p[i];
  }

//————————————————二分探索——————————————————————————
//部分列の組み合わせはn*(n+1)/2個
//mid以上の中央値が(n*(n+1)+3)/4個以上

ll ok = 0;
ll ng = 1000000001;
while (ng - ok > 1) {
  ll mid = (ng + ok) / 2;

  vector<ll> r=p;
  rep(i,n){
    if(r[i]>=mid) r[i]=1;
    else r[i]=-1;
  }
  vector<ll> s(n+1);
  s[0]=0;
  rep(i,n){
    s[i+1]=s[i]+r[i];
  }

  ll cnt=0;
  vector<ll> q=s;
  sort(all(q));
  uniq(q);

  BIT<ll> bit(n+10);
  rep(i,n+1){
    ll id=lb(q,s[i]);
    cnt += (ll)bit.sum0(id);
    bit.add0(id,1ll);
  }

  (cnt>=(ll)(n*(ll)(n+1)+3ll)/4ll? ok : ng) = mid;
}

  cout << ok << endl;

}


Measured memory_bytes: 7184