#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define MAXN 100001
#define LL long long
using namespace std;
LL n,a[MAXN],s[MAXN],l,r,f[MAXN*2+10],b[MAXN],c[MAXN*2+10];
void add(LL x)
{
	while(x<=2*MAXN) f[x]++,x+=x&-x;
	return ;
}
LL query(LL x)
{
	LL tot=0;
	while(x>0) tot+=f[x],x-=x&-x;
	return tot;
}
bool check(LL x)
{
	memset(f,0,sizeof f);
	for(LL i=1;i<=n;i++)
	{
		if(a[i]>=x) c[i]=1;
		else c[i]=-1;
		c[i]+=c[i-1];
	}
	LL ans=0;
	for(LL i=0;i<=n;i++) 
	ans+=(query(c[i]+MAXN)),add(c[i]+MAXN);
	//for(LL i=2;i<=n;i++) if
	//ans=n*(n-1)/2-ans;
	return ans>=n*(n+1)/4?true:false;
}
void slove()
{
	l=1,r=n;LL mid;
	while(l<=r)
	{
		mid=l+r>>1;
		if(check(s[mid])) l=mid+1;
		else r=mid-1;
	}
	cout<<s[r];
	return;
}
int main()
{
	cin>>n;
	for(LL i=1;i<=n;i++) cin>>a[i],s[i]=a[i];
	sort(s+1,s+n+1);
	slove();
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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>
#include<iostream>
#include<cstring>
#include<algorithm>
#define MAXN 100001
#define LL long long
using namespace std;
LL n,a[MAXN],s[MAXN],l,r,f[MAXN*2+10],b[MAXN],c[MAXN*2+10];
void add(LL x)
{
	while(x<=2*MAXN) f[x]++,x+=x&-x;
	return ;
}
LL query(LL x)
{
	LL tot=0;
	while(x>0) tot+=f[x],x-=x&-x;
	return tot;
}
bool check(LL x)
{
	memset(f,0,sizeof f);
	for(LL i=1;i<=n;i++)
	{
		if(a[i]>=x) c[i]=1;
		else c[i]=-1;
		c[i]+=c[i-1];
	}
	LL ans=0;
	for(LL i=0;i<=n;i++) 
	ans+=(query(c[i]+MAXN)),add(c[i]+MAXN);
	//for(LL i=2;i<=n;i++) if
	//ans=n*(n-1)/2-ans;
	return ans>=n*(n+1)/4?true:false;
}
void slove()
{
	l=1,r=n;LL mid;
	while(l<=r)
	{
		mid=l+r>>1;
		if(check(s[mid])) l=mid+1;
		else r=mid-1;
	}
	cout<<s[r];
	return;
}
int main()
{
	cin>>n;
	for(LL i=1;i<=n;i++) cin>>a[i],s[i]=a[i];
	sort(s+1,s+n+1);
	slove();
	return 0;
}
```