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 <cstring>
using namespace std;
const int maxn=1e6+10;
int a[maxn],b[maxn];
struct node{
	int val,loc;
}sub[maxn];
long long tre[maxn],n;
int cmp(const node &a,const node &b){
	if(a.val==b.val) return a.loc<b.loc;
	else return a.val<b.val;
}
inline int lowbit(int x){
	return x&(-x);
}
void add(int pos,int val){
	while(pos<=n){
		tre[pos]+=val;
		pos+=lowbit(pos);
	}
	return;
}
long long query(int pos){
	long long sum=0;
	while(pos>0){
		sum+=tre[pos];
		pos-=lowbit(pos);
	}
	return sum;
}
long long check(int val){
	memset(tre,0,sizeof(tre));
	long long ans=0;
	for(int i=1;i<=n;i++){
		if(a[i]<val) sub[i].val=-1;
		else sub[i].val=1;
		sub[i].val+=sub[i-1].val;
		sub[i].loc=i;
	}
	sort(sub+1,sub+n+1,cmp);
	for(int i=1;i<=n;i++){
		add(sub[i].loc,1);
		ans+=query(sub[i].loc-1);
		if(sub[i].val>=0) ans++;
	}
	return ans;
}
int erf(int l,int r){
	int mid;
	long long _std=(n+1)*n/4+(((n+1)*n/2)&1);
	while(l<r){
		mid=(l+r+1)>>1;
		if(check(b[mid])>=_std) l=mid;
		else r=mid-1;
	}
	return l;
}
int main(){
	cin>>n;
	for(int i=1;i<=n;i++) {
		scanf("%d",&a[i]);b[i]=a[i];
	}
	
	sort(b+1,b+n+1);
	cout<<b[erf(1,n)];
	
	return 0;
}
