#include <bits/stdc++.h>
#define lowbit(x) x&-x
using namespace std;

typedef long long ll;
static const int inf = 0x7fffffff;
static const int N = 1e5 + 10;
int n, a[N], cnt[N*10], pre[N*10];

void add (int x)
{
	for (int i = x; i <= N << 1; i += lowbit(i))
		++cnt[i];
}

ll query (int x)
{
	ll sum = 0;
	for (int i = x; i > 0; i -= lowbit(i))
		sum += cnt[i];
	return sum;
}

bool check (int x)
{
	for (int i = 1; i <= N << 1; ++i)
		cnt[i] = 0;
	pre[0] = 0;
	for (int i = 1; i <= n; ++i)
		pre[i] = pre[i-1] + (a[i] >= x ? 1 : -1);
	ll sum = 0;
	for (int i = 0; i <= n; ++i)
	{
		sum += query(pre[i] + N);
		add(pre[i]+N);
	}
	return sum >= 1ll * n * (n + 1) >> 2;
}

int main ()
{
	cin >> n;
	int l = 0, r = 0;
	for (int i = 1; i <= n; ++i)
		cin >> a[i], r = max(r, a[i]);
	int ans = 0;
	while (l <= r)
	{
		int mid = (l + r) >> 1;
		if (check(mid))
			l = mid + 1;
		else
			r = mid - 1;
	}
	cout << r;
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
#define lowbit(x) x&-x
using namespace std;

typedef long long ll;
static const int inf = 0x7fffffff;
static const int N = 1e5 + 10;
int n, a[N], cnt[N*10], pre[N*10];

void add (int x)
{
	for (int i = x; i <= N << 1; i += lowbit(i))
		++cnt[i];
}

ll query (int x)
{
	ll sum = 0;
	for (int i = x; i > 0; i -= lowbit(i))
		sum += cnt[i];
	return sum;
}

bool check (int x)
{
	for (int i = 1; i <= N << 1; ++i)
		cnt[i] = 0;
	pre[0] = 0;
	for (int i = 1; i <= n; ++i)
		pre[i] = pre[i-1] + (a[i] >= x ? 1 : -1);
	ll sum = 0;
	for (int i = 0; i <= n; ++i)
	{
		sum += query(pre[i] + N);
		add(pre[i]+N);
	}
	return sum >= 1ll * n * (n + 1) >> 2;
}

int main ()
{
	cin >> n;
	int l = 0, r = 0;
	for (int i = 1; i <= n; ++i)
		cin >> a[i], r = max(r, a[i]);
	int ans = 0;
	while (l <= r)
	{
		int mid = (l + r) >> 1;
		if (check(mid))
			l = mid + 1;
		else
			r = mid - 1;
	}
	cout << r;
	return 0;
}
```