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;
int n,a[100005],bit[200005];
int b[200005];
void add(int x){
	while (x<200005){
		bit[x]++;
		x+=(x&-x);
	}
}
int num(int x){
	int tmp=0;
	while(x){
		tmp+=bit[x];
		x-=x&-x;
	}
	return tmp;
}
bool check(int x){
	memset(bit,0,sizeof(bit));
	for(int i=0;i<n;i++)b[i]=(a[i]<=x)-(a[i]>x);
	long long all=0,cnt=0;
	for(int i=0;i<n;i++){
		all+=b[i];
		add(all+100000);
		cnt+=(all>0);
		cnt+=num(all+99999);//Ö»Òªall-Ä³¸öÇø¼ä>=0 ¼´¿É³ÉÁ¢ 
	//	cout<<cnt<<' ';
	//	for(int i=-3;i<=3;i++)cout<<bit[i+100000];
	//	cout<<endl;
	}
	//cout<<endl;
	return cnt>=((n*1ll*(n+1))/2)/2+1;
}
int main(){
	cin>>n;
	vector<int>v;
	for(int i=0;i<n;i++)cin>>a[i],v.push_back(a[i]);
	//sort(a,a+n);
	sort(v.begin(),v.end());
	v.erase(unique(v.begin(), v.end()), v.end());
    int l=0,r=v.size()-1;
	while(l<r){
		int mid=(l+r)/2;
		if(check(v[mid]))
			r=mid;
		else
			l=mid+1;
	}
	cout<<v[l];
}