#include<bits/stdc++.h>
using namespace std;
int a[100010],d[100010],b[100010],n;
long long t,tt;
void sol(int l,int r)
{
	if(l>=r) return ;
	int mid=(l+r)/2,p,q,x,i;
	sol(l,mid);
	sol(mid+1,r);
	p=l;
	q=mid+1;
	x=l;
	while(p<=mid&&q<=r)
	{
		if(d[p]<d[q]) b[x++]=d[p++];
		else
		{
			t+=(p-l);
			b[x++]=d[q++];
		}
	}
	while(p<=mid) b[x++]=d[p++];
	while(q<=r)
	{
		t+=(p-l);
		b[x++]=d[q++];
	}
	for(i=l; i<=r; i++)
		d[i]=b[i];
}
bool check(int x)
{
	t=0;
	int i;
	for(i=1; i<=n; i++)
		if(a[i]<=x) d[i]=1;
		else d[i]=-1;
	for(i=1; i<=n; i++)
		d[i]+=d[i-1];
	sol(0,n);
	return t>=(tt/2+1);
}
int main()
{
	int i,l=1e9,r=-1e9,mid;
	tt=0;
	scanf("%d",&n);
	for(i=1; i<=n; i++)
	{
		scanf("%d",&a[i]);
		l=min(l,a[i]);
		r=max(r,a[i]);
		tt+=i;
	}
	while(l<r)
	{
		mid=(l+r)/2;
		if(check(mid)) r=mid;
		else l=mid+1;
	}
	printf("%d",r);
	return 0;
}
### Prompt

Please create a solution in Cpp 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;
int a[100010],d[100010],b[100010],n;
long long t,tt;
void sol(int l,int r)
{
	if(l>=r) return ;
	int mid=(l+r)/2,p,q,x,i;
	sol(l,mid);
	sol(mid+1,r);
	p=l;
	q=mid+1;
	x=l;
	while(p<=mid&&q<=r)
	{
		if(d[p]<d[q]) b[x++]=d[p++];
		else
		{
			t+=(p-l);
			b[x++]=d[q++];
		}
	}
	while(p<=mid) b[x++]=d[p++];
	while(q<=r)
	{
		t+=(p-l);
		b[x++]=d[q++];
	}
	for(i=l; i<=r; i++)
		d[i]=b[i];
}
bool check(int x)
{
	t=0;
	int i;
	for(i=1; i<=n; i++)
		if(a[i]<=x) d[i]=1;
		else d[i]=-1;
	for(i=1; i<=n; i++)
		d[i]+=d[i-1];
	sol(0,n);
	return t>=(tt/2+1);
}
int main()
{
	int i,l=1e9,r=-1e9,mid;
	tt=0;
	scanf("%d",&n);
	for(i=1; i<=n; i++)
	{
		scanf("%d",&a[i]);
		l=min(l,a[i]);
		r=max(r,a[i]);
		tt+=i;
	}
	while(l<r)
	{
		mid=(l+r)/2;
		if(check(mid)) r=mid;
		else l=mid+1;
	}
	printf("%d",r);
	return 0;
}
```