#include <iostream>
#include <algorithm>
using namespace std;

const int MAX = 100000;

class BIT {
public:
	BIT(int n) {
		size = n + 1;
		bit = new int[size];

		for (int i = 0; i < size; i++)bit[i] = 0;
	}
	~BIT() {
		delete[] bit;
	}

	void add(int i, int n) {
		i += 1;
		while (i < size) {
			bit[i] += n;
			i += i & -i;
		}
	}

	int sum(int i) {
		i += 1;
		int ret = 0;
		while (i) {
			ret += bit[i];
			i -= i & -i;
		}
		return ret;
	}
	int size;
	int* bit;
};

int N;
int a[MAX];

bool solve(int x) {
	long long cnt = 0;
	int sum = 0;
	BIT bit(N * 2);

	for (int i = 0; i < N; i++) {
		sum += a[i] >= x ? 1 : -1;
		cnt += bit.sum(N + sum) + (sum >= 0 ? 1 : 0);
		bit.add(N + sum, 1);
	}

	if (cnt >= ((long long)N * (N + 1) / 2) / 2)return true;
	return false;
}

int main() {
	cin >> N;
	int begin = 0;
	int end = 0;
	int ans = 0;

	for (int i = 0; i < N; i++) {
		cin >> a[i];
		end = max(end, a[i] + 1);
	}

	if (N == 1)ans = a[0];
	else {
		while (begin < end) {
			int mid = (begin + end) / 2;

			if (solve(mid)) {
				ans = mid;
				begin = mid + 1;
			}
			else {
				end = mid;
			}
		}
	}

	cout << ans << endl;

	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <iostream>
#include <algorithm>
using namespace std;

const int MAX = 100000;

class BIT {
public:
	BIT(int n) {
		size = n + 1;
		bit = new int[size];

		for (int i = 0; i < size; i++)bit[i] = 0;
	}
	~BIT() {
		delete[] bit;
	}

	void add(int i, int n) {
		i += 1;
		while (i < size) {
			bit[i] += n;
			i += i & -i;
		}
	}

	int sum(int i) {
		i += 1;
		int ret = 0;
		while (i) {
			ret += bit[i];
			i -= i & -i;
		}
		return ret;
	}
	int size;
	int* bit;
};

int N;
int a[MAX];

bool solve(int x) {
	long long cnt = 0;
	int sum = 0;
	BIT bit(N * 2);

	for (int i = 0; i < N; i++) {
		sum += a[i] >= x ? 1 : -1;
		cnt += bit.sum(N + sum) + (sum >= 0 ? 1 : 0);
		bit.add(N + sum, 1);
	}

	if (cnt >= ((long long)N * (N + 1) / 2) / 2)return true;
	return false;
}

int main() {
	cin >> N;
	int begin = 0;
	int end = 0;
	int ans = 0;

	for (int i = 0; i < N; i++) {
		cin >> a[i];
		end = max(end, a[i] + 1);
	}

	if (N == 1)ans = a[0];
	else {
		while (begin < end) {
			int mid = (begin + end) / 2;

			if (solve(mid)) {
				ans = mid;
				begin = mid + 1;
			}
			else {
				end = mid;
			}
		}
	}

	cout << ans << endl;

	return 0;
}
```