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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstring>

using namespace std;

inline int gi()
{
	int f = 1, x = 0;
	char c = getchar();
	while (c < '0' || c > '9')
	{
		if (c == '-') f = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9')
	{
		x = x * 10 + c - '0';
		c = getchar();
	}
	return f * x;
}

int n, a[100005], s[1000005], t[1000005];

inline void add(int d)
{
	for (; d <= 200010; d = d + (d & (-d))) ++t[d];
}

inline long long getans(int y)
{
	long long sum = 0;
	for (; y; y = y - (y & (-y))) sum = sum + t[y];
	return sum;
}	 

inline bool check(int x)
{
	memset(t, 0, sizeof(t));
	s[0] = 0;
	for (int i = 1; i <= n; i++)
	{
		s[i] = s[i - 1] + ((a[i] >= x) ? (1) : (-1));
	}
	long long sum = 0;
	for (int i = 0; i <= n; i++)
	{
		sum = sum + getans(s[i] + 100005);
		add(s[i] + 100005);
	}
	return sum >= (1ll * n * (n + 1) / 4);
}

int main()
{
	n = gi();
	int l = 0, r = 0;
	for (int i = 1; i <= n; i++) a[i] = gi(), r = max(r, a[i]);
	while (l <= r)
	{
		int mid = (l + r) >> 1;
		if (check(mid)) 
		{
			l = mid + 1;
		}
		else r = mid - 1;
	}
	printf("%d\n", r);
	return 0;
}