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;
const int MAXn = 1e5 + 10, INF = 1e9;

long long last[2 * MAXn], cnt[2 * MAXn], n, ar[MAXn], psum[MAXn];

bool check(int a) {
	fill_n(last, 2 * MAXn, -1);
	fill_n(cnt, 2 * MAXn, 0);
	fill_n(psum, MAXn, 0);
	long long psum1 = 0, ans = 0;
	for (int i = 0; i < n; i++) {
		if (ar[i] > a)
			psum1--;
		else
			psum1++;
		int hp = last[psum1 + MAXn] + 1;
		
		if (psum[hp] < psum1) {
			cnt[psum1 + MAXn] += (i - hp);
		}
		ans += cnt[psum1 + MAXn] + (psum1 > 0);
		last[psum1 + MAXn] = i;
		psum[i] = psum1;
	}
//	cout << ans << endl;
	long long hp2 = (n * (n + 1)) / 2;
	return hp2 / 2 < ans;
}
int main() {
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> ar[i];
	if (n == 1)
		return cout << ar[0],0;
	
	int st = 0, en = INF + 10;
	while(st + 1 < en) {
		int med = (st + en) / 2;
		if (check(med))
			en = med;
		else
			st = med;
	}
	cout << en;
}