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<vector>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
int n;
int a[100000+5];
int c[100000+5];
int tree[3*100000+5];

int lowbit(int x){
	return x&-x;
}

void update(int x){
	while(x<=200010){
		tree[x]++;
		x+=lowbit(x);
	}
}

ll query(int x){
	ll res=0;
	while(x>0){
		res+=tree[x];
		x-=lowbit(x);
	}
	return res;
}

bool check(int x){
	c[0]=0;
	for(int i=1;i<=n;i++) c[i]=c[i-1]+(a[i-1]>=x ? 1:-1);
	memset(tree,0,sizeof(tree));
	ll sum=0;
	for(int i=0;i<=n;i++){
		sum+=query(c[i]+100005);
		update(c[i]+100005);
	}
	return sum >= 1ll*n*(n+1)/4;
}

int main(){
	cin >> n;
	int l=0,r=-1e9;
	for(int i=0;i<n;i++){
		cin >> a[i];
		r=max(a[i],r);
	}
	while(l<=r){
		int mid=(l+r)>>1;
		if(check(mid)) l=mid+1;
		else r=mid-1;
	}
	cout << r << endl;
	return 0;
}
