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;
const int N = 1e5 + 100, inF = 1e9 + 100;

ll n, a[N], ps[N];
unordered_map<int, int> cnt;

ll check(ll num) {
	ll res = 0, now = 0;
	cnt.clear();
	cnt[0] = 1;
	for (int i = 0; i < n; i++) {
		ps[i] = (i? ps[i - 1]: 0) + (a[i] <= num? 1: -1);
		if(a[i] <= num)
			now += cnt[ps[i] - 1];
		else
			now -= cnt[ps[i]];
		cnt[ps[i]]++;
		res += now;
	}
	return res;
}

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	ll want = (n + 1) * n / 4 + 1;
	ll l = 0, r = inF;
	while(r - l > 1) {
		ll c = (l + r) >> 1;
		if(check(c) >= want)
			r = c;
		else
			l = c;
	}
	cout << r;
	return 0;
}