#include <cstdio>
using namespace std;
const int offset = 100010;

int N, a[100010], bit[200020];

void add(int i)
{
	for (; i < 200020; i += i&-i)
		bit[i]++;
}

int sum(int i)
{
	int ret = 0;
	for (; i > 0; i -= i&-i)
		ret += bit[i];
	return ret;
}

bool check(int med)
{
	long long total = 0;
	int s = 0, i;
	for (i = 0; i < 200020; i++)
		bit[i] = 0;

	for (i = 1; i <= N; i++)
	{
		add(s + offset);
		s += (a[i] >= med) ? 1 : -1;
		total += sum(s + offset);
	}

	return total*4 >= (long long)N*(N+1);
}

int main()
{
	int max = 0;
	scanf("%d", &N);
	for (int i = 1; i <= N; i++) {
		scanf("%d", &a[i]);
		if (max < a[i])
			max = a[i];
	}

	int l = 1, r = max+1, med;
	while (r-l > 1)
	{
		med = (l+r)/2;
		if (check(med))
			l = med;
		else
			r = med;
	}

	printf("%d\n", l);
}
### Prompt

Your task is to create a CPP solution to 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 <cstdio>
using namespace std;
const int offset = 100010;

int N, a[100010], bit[200020];

void add(int i)
{
	for (; i < 200020; i += i&-i)
		bit[i]++;
}

int sum(int i)
{
	int ret = 0;
	for (; i > 0; i -= i&-i)
		ret += bit[i];
	return ret;
}

bool check(int med)
{
	long long total = 0;
	int s = 0, i;
	for (i = 0; i < 200020; i++)
		bit[i] = 0;

	for (i = 1; i <= N; i++)
	{
		add(s + offset);
		s += (a[i] >= med) ? 1 : -1;
		total += sum(s + offset);
	}

	return total*4 >= (long long)N*(N+1);
}

int main()
{
	int max = 0;
	scanf("%d", &N);
	for (int i = 1; i <= N; i++) {
		scanf("%d", &a[i]);
		if (max < a[i])
			max = a[i];
	}

	int l = 1, r = max+1, med;
	while (r-l > 1)
	{
		med = (l+r)/2;
		if (check(med))
			l = med;
		else
			r = med;
	}

	printf("%d\n", l);
}
```