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

#define int long long

int n,sz;
int a[100001], b[100001], p[100001];
int BIT[200001];

void update(int x)
{
	if (x == 0) x = 1;
	for(;x<=200000;x+=x&(-x))	
		BIT[x]++;
	return;
}

int query(int x)
{
	int ans = 0;
	for(;x>0;x-=x&(-x))
		ans += BIT[x];
	return ans;
}

bool check(int x)
{
	for (int i=1;i<=n;i++)
	{
		if (a[i] >= x)
			b[i] = 1;
		else
			b[i] = 0;
		p[i] = p[i-1] + b[i];
	}
	for (int i=0;i<=200000;i++)
		BIT[i] = 0;
	update(100000);
	int cur = 0;
	for (int i=1;i<=n;i++)
	{
		cur += query(2*p[i] - i + 100000);
		update(2*p[i] - i + 100000);
	}
	return (cur >= sz);
}

int32_t main() 
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin >> n;
	for (int i=1;i<=n;i++)
		cin >> a[i];
	sz = n*(n+1)/2;
	sz = (sz+1)/2;
	int l = 1, r = 1e9;
	int res;
	while(l <= r)
	{
		int mi = (l+r)/2;
		if (check(mi))
		{
			res = mi;
			l = mi + 1;
		}
		else
			r = mi - 1;
	}
	cout << res;
	return 0;
}
### Prompt

Please formulate 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 <bits/stdc++.h>
using namespace std;

#define int long long

int n,sz;
int a[100001], b[100001], p[100001];
int BIT[200001];

void update(int x)
{
	if (x == 0) x = 1;
	for(;x<=200000;x+=x&(-x))	
		BIT[x]++;
	return;
}

int query(int x)
{
	int ans = 0;
	for(;x>0;x-=x&(-x))
		ans += BIT[x];
	return ans;
}

bool check(int x)
{
	for (int i=1;i<=n;i++)
	{
		if (a[i] >= x)
			b[i] = 1;
		else
			b[i] = 0;
		p[i] = p[i-1] + b[i];
	}
	for (int i=0;i<=200000;i++)
		BIT[i] = 0;
	update(100000);
	int cur = 0;
	for (int i=1;i<=n;i++)
	{
		cur += query(2*p[i] - i + 100000);
		update(2*p[i] - i + 100000);
	}
	return (cur >= sz);
}

int32_t main() 
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin >> n;
	for (int i=1;i<=n;i++)
		cin >> a[i];
	sz = n*(n+1)/2;
	sz = (sz+1)/2;
	int l = 1, r = 1e9;
	int res;
	while(l <= r)
	{
		int mi = (l+r)/2;
		if (check(mi))
		{
			res = mi;
			l = mi + 1;
		}
		else
			r = mi - 1;
	}
	cout << res;
	return 0;
}
```