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 ull = unsigned long long;
using pii = pair<int,int>; 
#define INF INT_MAX //(1<<(4*4-1))
#define MOD 1000000007
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()

struct BIT{
   int sz;
   vector<ll>nd;

   BIT(int n):nd(vector<ll>(n+1,0)),sz(n){}

   ll sum(int i){
      if(!i) return 0;
      return nd[i]+sum(i-(i&-i));
   }

   void add(ll i,int x){
      if(i>sz)return;
      nd[i]+=x;
      add(i+(i&-i),x);
   }

};

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

   int N;
   cin>>N;
   vector<int>a(N);
   for(int i=0;i<N;i++)cin>>a[i];

   int offset = 100001;
   int A = *max_element(rng(a));
   ll ok=0;
   ll ng=INF;
   ll md;
   while(ok+1!=ng){
      md=(ok+ng)/2;
      ll sum=0;
      BIT bit(201010);
      int tot=0;
      bit.add(offset,1); //0
      for(int i=0;i<N;i++){
         if(md<=a[i])tot++;
         else tot--;
         sum+=bit.sum(tot+offset);
         bit.add(tot+offset,1);
      }
      if(sum>=((1LL*N*(N-1)/2+N+1)/2)) ok=md;
      else ng=md;
   }
   cout<<ok<<endl;
   return 0;
}

