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>
typedef long long ll;
using namespace std;

vector<int> A;
ll N;
vector<int> v,w;

ll merge_count(vector<int> &a){
	int n=a.size();
	if(n<=1) return 0;

	ll cnt=0;
	vector<int> b(a.begin(),a.begin()+n/2);
	vector<int> c(a.begin()+n/2,a.end());

	cnt+=merge_count(b);
	cnt+=merge_count(c);
	//この段階でbcはソート済み

	int ai=0,bi=0,ci=0;
	while(ai<n){
		if(bi<b.size() && (ci==c.size()||b[bi]<c[ci])){
			a[ai++]=b[bi++];
	}
		else {
			cnt+=n/2-bi;
			a[ai++]=c[ci++];
		}
	}
	return cnt;
}


bool flag(int x,vector<int> v){
  int i;
  int l[N];
  for(i=0;i<N;i++) l[i]=2*(v[i]>=x)-1;
  vector<int> s;
  s.push_back(0);
  for(i=1;i<=N;i++) s.push_back(s[i-1]+l[i-1]);
  reverse(s.begin(),s.end());
  if (merge_count(s)>=(N*(N+1)/2+1)/2) return 1;
  else return 0;
}
	


int main(){
  int i,j;
  int x;
  cin>>N;
  for(i=0;i<N;i++) {
	cin>>x;
	v.push_back(x);
	w.push_back(x);
  }
  
  sort(w.begin(),w.end());
  
  int l=0,r=N-1;
  while(r-l>=1){
	int m=(l+r)/2;
	if(flag(w[m],v)) {
		l=m+1;
	}
	else {
		r=m;
	}
  }
  if(flag(w[r],v)) cout<<w[r]<<endl;
  else cout<<w[r-1]<<endl;
  return 0;
}