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>

#define ll long long
#define mx  200005
#define mod 1000000009
#define lim 10000000
#define inf 2000000000
#define N 18
#define eps 100000
using namespace std;
int fre[mx],sum[mx],a[mx];
ll n;

ll getCount(int x)
{
	memset(fre,0,sizeof(fre));

	sum[0]=0;
	for(int i=1;i<=n;i++)
	{
		if(a[i]>=x)sum[i]=sum[i-1]+1;
		else sum[i]=sum[i-1]-1;
	}

	fre[eps]=1;
	int pre=0;
	ll ans=0;

	for(int i=1;i<=n;i++)
	{
		if(sum[i]>sum[i-1])pre+=fre[sum[i]+eps]+1;
		else pre=pre-fre[sum[i-1]+eps]+1;
		fre[sum[i]+eps]++;
		ans+=1LL*pre;
	}
	return ans;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin>>n;
	ll req=n*(n+1)/2;
	req=ceil(req/2);
	vector<int>v;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		v.push_back(a[i]);
	}
	sort(v.begin(),v.end());
	v.erase(unique(v.begin(),v.end()),v.end());
	int lo=0,hi=v.size()-1;

	for(int i=1;i<=30;i++)
	{
		int mid=(lo+hi+1)/2;
		if(getCount(v[mid])>=req)
			lo=mid;
		else
			hi=mid-1;
	}

	cout<<v[lo]<<"\n";
	return 0;
}
