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<algorithm>
#include<cstring>
#define ll long long
using namespace std;
const int maxn=100050;
int n;
int a[maxn],pre[maxn];
ll tr[maxn*10];
inline int lowbit(int x){
	return x&(-x);
}
inline void add(int x){
	for(register int i=x;i<=2*maxn;i+=lowbit(i)){
		tr[i]++;
	}
}
inline ll query(int x){
	ll ans=0;
	for(register int i=x;i>0;i-=lowbit(i)){
		ans+=tr[i];
	}
	return ans;
}
inline bool judge(int x){
	for(register int i=0;i<=2*maxn;i++)tr[i]=0;
	pre[0]=0;
	for(register int i=1;i<=n;i++){
		pre[i]=pre[i-1];
		if(a[i]>=x)pre[i]++;
		else pre[i]--;
	}
	ll cnt=0;
	for(register int i=0;i<=n;i++){
		cnt+=query(pre[i]+maxn);
		add(pre[i]+maxn);
	}
	if(cnt>=(ll)n*(n+1)/4)return 1;
	return 0;
}
int main(){
	int l=0,r=0;
	scanf("%d",&n);
	for(register int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		r=max(a[i],r);
	}
	while(l<=r){
		int m=(l+r)>>1;
		if(judge(m))l=m+1;
		else r=m-1;
	}
	cout<<r<<endl;
}