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 maxn = 2e5 + 10;
const int N = 1e5 + 1;

int a[maxn], par[maxn], fen[maxn];
int n;

int get(int x){
	int ret = 0;
	for (x += N; x; x -= x & -x)
		ret += fen[x];
	return ret;
}

void add(int x, int val){
	for (x += N; x < maxn; x += x & -x)
		fen[x] += val;
}

ll check(int x){
	for (int i = 1; i <= n; i++)
		par[i] = (a[i] < x ? 1 : -1);
	for (int i = 1; i <= n; i++)
		par[i] += par[i - 1];

	add(par[0], +1);
	ll ret = 0;
	for (int i = 1; i <= n; i++) {
		ret += get(par[i] - 1);
		add(par[i], +1);
	}
	// vitex
	for (int i = 0; i <= n; i++)
		add(par[i], -1);
	return ret;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	ll tot = 1ll * n * (n + 1) / 2;
	int lo = 1, hi = 1000 * 1000 * 1000 + 1;
	while (hi - lo > 1){
		int mid = (lo + hi) / 2;
		if (check(mid) > tot / 2)
			hi = mid;
		else
			lo = mid;
	}
	cout << lo << endl;
}