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<algorithm>
#include<cstring>
#define ll long long 
ll x;
int t[200050],n,a[100050],b[100050],u[100050];
void add(int x){for (int i=x;i<=200000;i+=i&(-i))t[i]++;}
int query(int x){int ret=0;for (int i=x;i;i-=i&(-i))ret+=t[i];return ret;}
bool check(int p){
	memset(t,0,sizeof t);
	ll ret=0;
	for (int i=1;i<=n;i++)b[i]=(u[i]>=p)?1:-1;
	add(n);
	for (int i=1;i<=n;i++){
		b[i]=b[i]+b[i-1];
		ret+=query(n+b[i]);
		add(n+b[i]);
	} 
	return ret>=x;
}
int read(){
	int c=getchar(),ret=0;
	while (c<'0'||c>'9')c=getchar();
	while (c>='0'&&c<='9')ret=(ret<<1)+(ret<<3)+c-'0',c=getchar();
	return ret;
}
int main(){
	n=read();
	for (int i=1;i<=n;i++)u[i]=a[i]=read();
	x=(n*1ll*(n+1)/2+1)/2;
	std::sort(a+1,a+n+1);
	int tmp=std::unique(a+1,a+n+1)-a-1;
	int l=1,r=tmp,ans=1;
	while (l<=r){
		int mid=(l+r)>>1;
		if (check(a[mid]))ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d",a[ans]);
}