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<cstring>
#include<algorithm>
using namespace std;
long long N,a1[100005],a2[100005],F[200005],pre[100005],l,r,mid;
void insert(long long n)
{
	for(;n<=N*2+1;n+=n&-n)
	F[n]+=1;
}
int find(long long n)
{
	int cnt=0;
	for(;n;n-=n&-n)
	cnt+=F[n];
	return cnt;
}
bool check(long long n)
{
	memset(F,0,sizeof(F));
	memset(pre,0,sizeof(pre));
	long long cnt=0;
	long long zj=a2[n];
	for(int i=1;i<=N;i++)
	if(a1[i]<=zj) pre[i]=pre[i-1]-1 , cnt += pre[i] < 0 ? 1 : 0;
	else pre[i]=pre[i-1]+1;
	for(int i=1;i<=N;i++)
	{
		cnt+=i-find(pre[i]+N+1)-1;
		insert(pre[i]+N+1);
	}
	if(cnt<=N*(N+1)/2/2)
	return true;
	return false;
}
int main()
{
	scanf("%lld",&N);
	for(int i=1;i<=N;i++)
	{
		scanf("%lld",&a1[i]);
		a2[i]=a1[i];
	}
	sort(a2+1,a2+N+1);
	l=1,r=N;
	while(l+1<r)
	{
		int mid=(l+r)/2;
		if(check(mid))
		l=mid;
		else
		r=mid;
	}
	printf("%lld",a2[r]);
	return 0;
}