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
//Copyright(c)2018 Mstdream
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int N=200010;
int t[N],a[N],b[N],n;
void add(int x){
	for(;x<=2*n+1;x+=x&-x)t[x]++;
}
int qry(int x){
	int ret=0;
	for(;x;x-=x&-x)ret+=t[x];
	return ret;
}
bool check(int x){
	x++;
	LL ret=0;
	for(int i=1;i<=2*n;i++)t[i]=0;
	for(int i=1;i<=n;i++){
		b[i]=(a[i]>=x?1:-1);
		b[i]+=b[i-1];
		add(b[i-1]+n+1);
		ret+=qry(b[i]+n+1);
	}
	ret=(LL)n*(n+1)/2-ret;
	return ret>=((LL)n*(n+1)/2+2)/2;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	int l=1,r=1e9+1;
	while(l!=r){
		int mid=l+r>>1;
		if(check(mid))r=mid;
		else l=mid+1;
	}
	cout<<l<<endl;
}
