#include<bits/stdc++.h>
using namespace std;
const int N=100005;
int bit[N*2],a[N],s[N],n;
int sum(int i)
{
	int s=0;
	while(i)
	{
		s+=bit[i];
		i-=i&-i;
	}
	return s;
}
void add(int i,int x)
{
	while(i<=200000)
	{
		bit[i]+=x;
		i+=i&-i;	
	}	
}
int check(int x)
{
	memset(bit,0,sizeof(bit));
	s[0]=0;
	for(int i=1;i<=n;i++)
	{
		if(a[i]>=x)
			s[i]=s[i-1]+1;
		else
			s[i]=s[i-1]-1;
	}
	long long tot=0;
	for(int i=0;i<=n;i++)
	{
		tot+=sum(s[i]+100000);
		add(s[i]+100000,1);
	}
	if(tot>=1LL*n*(n+1)/4)
		return 1;
	else
		return 0;
}
int main()
{
	int lb=0,ub=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		ub=max(ub,a[i]+1);
	}
	while(ub-lb>1)
	{
		int mid=(lb+ub)>>1;
		if(check(mid))
			lb=mid;
		else
			ub=mid;
	}
	printf("%d\n",lb);
	return 0;
}

### Prompt

Generate 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;
const int N=100005;
int bit[N*2],a[N],s[N],n;
int sum(int i)
{
	int s=0;
	while(i)
	{
		s+=bit[i];
		i-=i&-i;
	}
	return s;
}
void add(int i,int x)
{
	while(i<=200000)
	{
		bit[i]+=x;
		i+=i&-i;	
	}	
}
int check(int x)
{
	memset(bit,0,sizeof(bit));
	s[0]=0;
	for(int i=1;i<=n;i++)
	{
		if(a[i]>=x)
			s[i]=s[i-1]+1;
		else
			s[i]=s[i-1]-1;
	}
	long long tot=0;
	for(int i=0;i<=n;i++)
	{
		tot+=sum(s[i]+100000);
		add(s[i]+100000,1);
	}
	if(tot>=1LL*n*(n+1)/4)
		return 1;
	else
		return 0;
}
int main()
{
	int lb=0,ub=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		ub=max(ub,a[i]+1);
	}
	while(ub-lb>1)
	{
		int mid=(lb+ub)>>1;
		if(check(mid))
			lb=mid;
		else
			ub=mid;
	}
	printf("%d\n",lb);
	return 0;
}

```