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<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<set>
using namespace std;
#define ll long long
const int N=100008;
ll n,nn,a[N],b[N+N];
ll add(ll x,ll y){while(x<=n+n){b[x]+=y;x+=x&(-x);}return 0;}
ll sum(ll x){ll y=0;while(x>0){y+=b[x];x-=x&(-x);}return y;}
ll pan(ll x)
{
	ll i,j,k;
	k=0;
	add(n+1,1);j=0;for(i=1;i<=n;i++){j+=(a[i]<=x);k+=sum(n+j+j-i);add(n+j+j-i+1,1);}
	add(n+1,-1);j=0;for(i=1;i<=n;i++){j+=(a[i]<=x);add(n+j+j-i+1,-1);}
	return k;
}
int main(void)
{
	std::ios::sync_with_stdio(false);
	ll i,left,right,mid;
	cin>>n;for(i=1;i<=n;i++)cin>>a[i];
	nn=(n+n*(n-1)/2)/2;
	left=1;right=1000000000;
	while(left+1<right)
	{
		mid=(left+right)>>1;
		if(pan(mid)>nn)right=mid;else left=mid+1;
	}
	if(pan(left)>nn)right=left;
	cout<<right;
	return 0;
}