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>
#include <cstdio>
#include <cstring>
#define int long long
#define N 100002
using namespace std;
int n,i,a[N],c[N*2],sum[N],l,r,mid,ans;
int lowbit(int x)
{
	return x&(-x);
}
void modify(int x,int y)
{
	for(int i=x;i<=N+n;i+=lowbit(i)) c[i]+=y;
}
int ask(int x)
{
	int ans=0;
	for(int i=x;i>=N-n;i-=lowbit(i)) ans+=c[i];
	return ans;
}
bool check(int x)
{
	int ans=0;
	memset(c,0,sizeof(c));
	for(int i=1;i<=n;i++){
		if(a[i]<x) sum[i]=sum[i-1]-1;
		else sum[i]=sum[i-1]+1;
	}
	for(int i=0;i<=n;i++){
		ans+=ask(sum[i]+N);
		modify(sum[i]+N,1);
	}
	if(ans>=n*(n+1)/4) return 1;
	return 0;
}
signed main()
{
	cin>>n;
	for(i=1;i<=n;i++){
		cin>>a[i];
		r=max(r,a[i]);
	}
	while(l<=r){
		mid=(l+r)/2;
		if(check(mid)){
			ans=mid;
			l=mid+1;
		}
		else r=mid-1;
	}
	cout<<ans<<endl;
	return 0;
}