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<cstring>
#include<cstdio>
#define ll long long
using namespace std;
const int N=100000+7;
int n,a[N],l,r,mid,c[3*N],t[N];
ll s[N];
int lowbit(int x){
	return x&-x;
}
void add(int x){
	for(int i=x;i<=2*N;i+=lowbit(i))
		c[i]++;
}
int query(int x){
	ll sum=0;
	for(int i=x;i>=1;i-=lowbit(i))
		sum+=c[i];
	return sum;
}
bool check(int now){
    for(int i=1;i<=2*N;i++)c[i]=0;		//用memset会出错 
	for(int i=1;i<=n;i++){
		s[i]=s[i-1]+(a[i]>=now ?1 :-1);
		//if(now==9)printf("%d ",s[i]);
	}
	//printf("\n");
	ll sum=0;
	for(int i=0;i<=n;i++){
		sum+=query(s[i]+N);
		add(s[i]+N);
	}
	//printf("%d %d\n",now,sum);
	return sum>=1ll*n*(n+1)/4;
}
int main(){
	//freopen("1.txt","r",stdin);
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		r=max(r,a[i]);
	}
	while(l<=r){			//一定要用 <=，不然会漏情况 
		mid=(l+r)>>1;
		if(check(mid))l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",r);
	return 0;
}
/*
3
10 30 20

*/
/*
10
5 9 5 9 8 9 3 5 4 3

*/