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<iostream>
#define int long long
using namespace std;
int a[1000001],n,ans;
int s[1000001],c[5000001];
int lowbit(int x)
{
	return x&(-x);
}
void add(int x,int d)
{
	for (;x<=n*4;x+=lowbit(x))
		c[x]+=d;
	return;
}
int sum(int x)
{
	ans=0;
	for (;x>=1;x-=lowbit(x))
		ans+=c[x];
	return ans;
}
bool check(int x)
{
	for (int i=1;i<=n*4;++i)
		c[i]=0;
	s[0]=0;
	int res=0;
	for (int i=1;i<=n;++i)
	{
		if (a[i]>=x)
			s[i]=s[i-1]+1;
		else
			s[i]=s[i-1]-1;
	}
	for (int i=0;i<=n;++i)
	{
		res+=sum(s[i]+n+1);
		add(s[i]+n+1,1);
	}
	return res>=(n+1)*n/4;
}
signed main()
{
	int first,last,mid,ans=-(1e+9),minn=1e+9,maxn=-(1e+9);
	cin>>n;
	for (int i=1;i<=n;++i)
	{
		cin>>a[i];
		maxn=max(maxn,a[i]);
		minn=min(minn,a[i]);
	}
	first=minn;
	last=maxn;
	ans=minn;
	while (first<=last)
	{
		mid=(first+last)/2;
		if (check(mid))
		{
			ans=max(ans,mid);
			first=mid+1;
		}
		else
			last=mid-1;
	}
	cout<<ans<<endl;
	return 0;
}