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;
const int offset=100000;
int n,a[100005],s[1000050],bit[1000050];
int lowb(int x)
{
	return x&(-x);
}
void add(int x)
{
	while(x<=2*offset)
	{
		bit[x]++;
		x+=lowb(x);
	}
}
long long cal(int x)
{
	long long ans=0;
	while(x>=1)
	{
		ans+=bit[x];
		x-=lowb(x);
	}
	return ans;
}
bool check(int x)
{
	memset(bit,0,sizeof(bit));
	for(int i=1;i<=n;i++)
		s[i]=s[i-1]+(a[i]>=x?1:-1);
	long long sum=0;
	for(int i=0;i<=n;i++)
	{
		sum+=cal(s[i]+offset);
		add(s[i]+offset);
	}
	return sum>=1ll*n*(n+1)/4;
}
int main()
{
	cin>>n;
	int l=0,r=0;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		r=max(r,a[i]);
	}
	while(l<=r)
	{
		int mid=(l+r)/2;
		if(check(mid)) l=mid+1;
		else r=mid-1;
	}
	cout<<r<<endl;
	return 0;
}
