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 <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;
ll n, a[100005], b[100005], p[100005], pp[100005];

ll cdq(ll l, ll r)
{
	if (l == r) return 0;
	ll mid = (l + r) / 2;
	ll ret = cdq(l, mid) + cdq(mid + 1, r);
	ll i = l;
	ll now = l - 1;
	for (ll j = mid + 1; j <= r; j++)	
	{
		while (p[i] <= p[j] && i <= mid)
		{
			pp[++now] = p[i];
			i++;
		}
		pp[++now] = p[j];
		if (i > mid) 
		{
			for (ll k = j + 1; k <= r; k++) pp[++now] = p[k];
			break;
		}
		ret += (mid - i + 1);
	}
	for (ll k = i; k <= mid; k++) pp[++now] = p[k];
	for (ll i = l; i <= r; i++) p[i] = pp[i];
	return ret;
}

ll erfen(ll l, ll r)
{
	memset(p, 0, sizeof(p));
	if (l == r) return l;
	ll mid = (l + r) / 2;
	ll rel = b[mid];
	for (ll i = 1; i <= n; i++)
	{
		p[i] = p[i - 1];
		if (a[i] > rel) p[i]++;
		else p[i]--;
	}
	ll num = cdq(0, n);
	if (num >= ((n + 1) * n / 2) / 2 + 1) return erfen(l, mid);
	else return erfen(mid + 1, r);
}

int main()
{
	cin >> n;
	for (ll i = 1; i <= n; i++) cin >> a[i], b[i] = a[i];
	sort (b + 1, b + n + 1);
	cout << b[erfen(1, n)] << endl;
	return 0;
}