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<cstdio>
#include<iostream>
#include<queue>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<cstdlib>
#include<stack>
#include<ctime>
#include<map>
#include<list>
using namespace std;
typedef long long ll;

ll a[100005],s[100005],tmp[100005];
ll n,ans;

ll msort(ll l,ll r)
{
	ll ttmp=0;
	if(r>l+1)
	{
		ll mid=(l+r)/2;
		ttmp=msort(l,mid)+msort(mid,r);
		for(ll i=l; i<r; i++)
		{
			tmp[i]=s[i];
		}
		ll tmp1=l,tmp2=mid,tmp3=l;
		while(tmp1<mid && tmp2<r)
		{
			if(tmp[tmp1]<=tmp[tmp2]) s[tmp3++]=tmp[tmp1++];
			else
			{
				s[tmp3++]=tmp[tmp2++];
				ttmp+=mid-tmp1;
			}
		}
		while(tmp1<mid) s[tmp3++]=tmp[tmp1++];
		while(tmp2<r) s[tmp3++]=tmp[tmp2++];
	}
	return ttmp;
}

bool chack(ll mid)
{
	s[0]=0;
	for(ll i=1; i<=n; i++)
	{
		if(a[i]>=mid) s[i]=s[i-1]+1;
		else s[i]=s[i-1]-1;
	}
	ll ttmp=msort(0,n+1);
	if(ans-ttmp>=(ans+1)/2) return true;
	else return false;
}

int main()
{
	scanf("%lld",&n);
	for(ll i=1; i<=n; i++) scanf("%lld",&a[i]);
	ll l=1,r=1000000010,mid;
	ans=1ll*(n+1)*n/2;
	while(l<r-1)
	{
		mid=(l+r)/2;
		if(chack(mid)) l=mid;
		else r=mid;
	}
	printf("%lld\n",l);
 	return 0;
}	