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 <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#define ll long long
#define INF 1000000007
using namespace std;

const int MAXN=1e5+5;

int n,l,r;
ll cnt,sum;
int a[MAXN],b[MAXN],c[MAXN];

void work(int begin,int end){
	if(begin==end)
		return;
	int mid=(begin+end)/2;
	work(begin,mid);
	work(mid+1,end);
	int i=begin,j=mid+1,k=begin;
	while(i<=mid && j<=end){
		if(b[i]<b[j])
			c[k++]=b[i++];
		else{
			cnt+=(i-begin);
			c[k++]=b[j++];
		}
	}
	while(i<=mid)
		c[k++]=b[i++];
	while(j<=end){
		cnt+=(i-begin);
		c[k++]=b[j++];
	}
	for(int l=begin;l<=end;l++)
		b[l]=c[l];
}

bool check(int m){
	cnt=0;
	for(int i=1;i<=n;i++){
		if(a[i]<=m)
			b[i]=1;
		else
			b[i]=-1;
	}
	for(int i=1;i<=n;i++){
		b[i]+=b[i-1];
	}
	work(0,n);
	if(cnt>=(sum/2+1))
		return true;
	return false;
}

int main(){
	scanf("%d",&n);
	l=INF;r=0;sum=0;
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		l=min(l,a[i]);
		r=max(r,a[i]);
		sum+=i;
	}
	while(l<r){
		int m=(l+r)/2;
		if(check(m)){
			r=m;
		}else{
			l=m+1;
		}
	}
	printf("%d\n",r);
	return 0;
}

