#include <bits/stdc++.h>
typedef long long ll; 
 
using namespace std;

template <class T>
struct BIT {
	vector<T> data;
	BIT(ll n) {
		data.assign(n, 0);
	}

	T sum (ll i) {
		T res = 0;
		while (i > 0) {
			res += data[i];
			i -= i & -i;
		}
		return res;
	}

	void add(ll i, T x) {
		while (i < data.size()) {
			data[i] += x;
			i += i & -i;
		}
	}
};

int main() {
	ll N;
	cin >> N;
	vector<ll> a(N + 1);
	vector<ll> x(N);
	for (int i = 1; i <= N; ++i) { 
		cin >> a[i];
		x[i - 1] = a[i];
	}
	sort(x.begin(), x.end());
	x.erase(unique(x.begin(), x.end()), x.end());

	ll Mlen = (1 + N) * N / 2;
	ll Mlenc = (Mlen + 1) / 2;

	int l = 0, r = x.size();
	while (r - l > 1) {
		int mid = (r + l) / 2;

		vector<ll> d(N + 1, 0);
		for (int i = 1; i <= N; ++i) {
			if (a[i] >= x[mid]) { d[i] += d[i - 1] + 1; }
			else { d[i] += d[i - 1] - 1; }
		}

		BIT<ll> bit(2 * N + 2);
		ll inv = 0;
		for (int i = 0; i < N + 1; ++i) {
			inv += bit.sum(2 * N + 1) - bit.sum(d[i] + N + 1);
			bit.add(d[i] + N + 1, 1);
		}

		ll wl = (1 + N) * N / 2 - inv;
		if (wl >= Mlenc) { l = mid; }
		else { r = mid; }
	}

	cout << x[l] << endl;
	return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
typedef long long ll; 
 
using namespace std;

template <class T>
struct BIT {
	vector<T> data;
	BIT(ll n) {
		data.assign(n, 0);
	}

	T sum (ll i) {
		T res = 0;
		while (i > 0) {
			res += data[i];
			i -= i & -i;
		}
		return res;
	}

	void add(ll i, T x) {
		while (i < data.size()) {
			data[i] += x;
			i += i & -i;
		}
	}
};

int main() {
	ll N;
	cin >> N;
	vector<ll> a(N + 1);
	vector<ll> x(N);
	for (int i = 1; i <= N; ++i) { 
		cin >> a[i];
		x[i - 1] = a[i];
	}
	sort(x.begin(), x.end());
	x.erase(unique(x.begin(), x.end()), x.end());

	ll Mlen = (1 + N) * N / 2;
	ll Mlenc = (Mlen + 1) / 2;

	int l = 0, r = x.size();
	while (r - l > 1) {
		int mid = (r + l) / 2;

		vector<ll> d(N + 1, 0);
		for (int i = 1; i <= N; ++i) {
			if (a[i] >= x[mid]) { d[i] += d[i - 1] + 1; }
			else { d[i] += d[i - 1] - 1; }
		}

		BIT<ll> bit(2 * N + 2);
		ll inv = 0;
		for (int i = 0; i < N + 1; ++i) {
			inv += bit.sum(2 * N + 1) - bit.sum(d[i] + N + 1);
			bit.add(d[i] + N + 1, 1);
		}

		ll wl = (1 + N) * N / 2 - inv;
		if (wl >= Mlenc) { l = mid; }
		else { r = mid; }
	}

	cout << x[l] << endl;
	return 0;
}

```