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;
#define all(V) V.begin(),V.end()
#define ll long long
#define MOD 1000000007

ll seg[524288];

ll rsq(int s, int t, int k = 1, int l = 0, int r = 262143) {
	if (t < l || r < s)return 0;
	if (s <= l && r <= t)return seg[k];
	ll lv = rsq(s, t, k * 2, l, (l + r) / 2);
	ll rv = rsq(s, t, k * 2 + 1, (l + r) / 2 + 1, r);
	return lv + rv;
}

void upd(ll x) {
	x += 262144;
	seg[x]++;
	while (x) {
		x /= 2;
		seg[x] = seg[x * 2] + seg[x * 2 + 1];
	}
}

int main() {
	for (int i = 0;i < 524288;i++)seg[i] = 0;
	ll N, L;
	cin >> N;
	L = N * (N + 1) / 2;
	vector<ll> A(N), AR(N);
	for (int i = 0;i < N;i++) {
		cin >> A[i];
		AR[i] = A[i];
	}
	sort(all(AR));

	ll ng = -1, ok = N - 1, T, x, y, s;
	vector<ll> B(N + 1, 0);
	while (ng + 1 != ok) {
		T = (ok + ng) / 2;
		x = AR[T];

		for (int i = 0;i < 524288;i++)seg[i] = 0;
		for (int i = 0;i < N;i++) {
			if (A[i] > x)B[i + 1] = -1;
			else B[i + 1] = 1;

			B[i + 1] += B[i];
		}

		s = 0;
		for (int i = 0;i <= N;i++) {
			upd(B[i] + 100005);
			s += rsq(0, B[i] + 100004);
		}
		if (s < L / 2 + 1)ng = T;
		else ok = T;
	}
	cout << AR[ok] << endl;
}