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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
ll n, a[100010], b[100010], bit[200010];

ll sum(int p){
	ll res = 0;
	for(int i=p; i>0; i-=i&-i) res += bit[i];
	return res;
}

void add(int x, int p){
	for(int i=p; i<=2*n+1; i+=i&-i) bit[i] += x;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n;
	ll l = 1e18, r = -1e18;
	for(int i=0; i<n; ++i){
		cin >> a[i];
		l = min(l, a[i]);
		r = max(r, a[i]);
	}
	++r;
	while(l+1 != r){
		ll m = (l + r) / 2, cnt = 0;
//		cout << l << " " << r << " " << m << "\n";
		for(int i=1; i<=n; ++i)
			b[i] = (a[i-1] >= m ? 1 : -1) + b[i-1];
		memset(bit, 0, sizeof(bit));
		add(1, n+1);
		for(int i=1; i<=n; ++i){
			cnt += sum(b[i]+n+1);
			add(1, b[i]+n+1);
		}
//		cout << cnt << "\n";
		if(cnt >= (n*(n+1)/2+1)/2) l = m;
		else r = m;
	}
	cout << l << "\n";
}