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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e17;

template <typename T>
class BIT {
  int n;
  vector<T> bit;
public:
  BIT(int _n) {
    n = _n;
    bit = vector<T>(n+1,(T)0);
  }
  T sum(int i) {
    T s = 0;
    while(i > 0){
    s += bit[i];
    i -= i & -i;
  }
  return s;
  }
  void add(int i, T x) {
    while(i <= n){
    bit[i] += x;
    i += i & -i;
    }
  }
};

int main(){
  ll n;
  cin >> n;
  ll a[n+1];
  rep(i,n) cin >> a[i+1];
  ll l = 0, r = INF;
  while(r - l > 1) {
    ll m = (l + r) / 2;
    vector<int> s(n+1,0);
    BIT<ll> odd(4*n), even(4*n);
    ll num = 0;
    even.add(2*n,1);
    for(ll i = 1; i <= n; i++) {
      s[i] = s[i-1];
      if(a[i] >= m) s[i]++;
      ll k = 2*s[i] - i + 2*n;
      if(i % 2) {
        num += odd.sum(k);
        num += even.sum(k-1);
      } else {
        num += odd.sum(k-1);
        num += even.sum(k);
      }
      if(i % 2) odd.add(k,1);
      else even.add(k,1);
    }
    if(num >= (n*(n+1)/2 +1)/2) l = m;
    else r = m;
  }
  cout << l << endl;
}
