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;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pii;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const ll N = (ll)1e5 + 9;
ll a[N];
map<ll,ll> cnt;

int main(){
  fastIO;
  ll n;
  cin >> n;
  for(ll i = 1; i <= n; i ++ ) cin >> a[i];
  ll l = 1;
  ll r = (ll)1e9 + 9;
  ll md;
  ll cur;
  ll ss = 1;
  ll sum;
  ll kk;
  while(l + 1 < r){
    md = (l + r) / 2ll;
    cnt.clear(); // unnesecary log but who cares?
    cnt[0] = 1;
    cur = 0;
    ss = 1;
    sum = 0;
    for(ll i = 1; i <= n; i ++ ){
      if(a[i] >= md) {
        cur ++ ;
        cnt[cur]++;
        ss += cnt[cur];
      }
      else{
        ss -= cnt[cur];
        cur--;
        cnt[cur]++;
        ss++;
      }
      sum += ss;
    }
    sum -= n;
    kk = n*(n+1)/2ll;
    if(sum >= (kk+1)/2){
      l = md;
    }
    else
      r = md;
  }
  cout << l;
  return 0;
}