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
// D.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>

using namespace std;

class BIT {
	std::vector<int> bit;
	int size;
public:
	BIT() { }
	BIT(int sz) { init(sz); }
	void init(int sz) {
		bit = std::vector<int>((size = sz) + 1);
	}
	int sum(int i) {
		int s = 0;
		while (i > 0) {
			s += bit[i];
			i -= i & -i;
		}
		return s;
	}
	void add(int i, int x) {
		while (i <= size) {
			bit[i] += x;
			i += i & -i;
		}
	}
};

int main(int argc, char* argv[]) {
#ifdef _MSC_VER
	while (true)
#endif
	{
		int n = -1;
		cin >> n;
		if (n <= 0) return 0;
		vector<int> a(n);
		for (int i = 0; i < n; ++i) {
			cin >> a[i];
		}
		int left = -1, right = 1e9;
		while (right - left > 1) {
			int mid = (right + left) / 2;
			BIT bit(n * 2 + 1);
			long long sum = n + 1, cnt = 0, need = (long long)n * (n + 1) / 4 + 1;
			bit.add(sum, 1);
			for (int i = 0; i < n; ++i) {
				sum += a[i] > mid ? 1 : -1;
				cnt += bit.sum(2 * n) - bit.sum(sum);
				bit.add(sum, 1);
			}
			if (cnt >= need) {
				right = mid;
			} else {
				left = mid;
			}
		}

		cout << right << endl;
	}
	return 0;
}
