#include<bits/stdc++.h>
#define N 100005
using namespace std;
int n;
int a[N],b[N];
int s[N];
long long tmp;
int c[N];
void gb(int l,int r)
{
	if(l==r)return;
	int mid=(l+r)/2,h1=l,h2=mid+1,cnt=l;
	gb(l,mid);
	gb(mid+1,r);
	while(h1<=mid&&h2<=r)
	{
		if(s[h1]<=s[h2]) c[cnt++]=s[h1++];
		else c[cnt++]=s[h2++],tmp+=(mid-h1+1);
	}
	for(int i=h1;i<=mid;++i)c[cnt++]=s[i];
	for(int i=h2;i<=r;++i)c[cnt++]=s[i];
	for(int i=l;i<=r;++i) s[i]=c[i];
	//cout<<l<<" "<<r<<" "<<tmp<<endl;
}
bool check(int x)
{
	memset(s,0,sizeof(s));
	for(int i=1;i<=n;++i)
		if(a[i]<x) s[i]=s[i-1]-1;
		else s[i]=s[i-1]+1;
	//for(int i=0;i<=n;++i)
	//cout<<s[i];
	//	cout<<endl;
	tmp=0;gb(0,n);
	//for(int i=0;i<=n;++i)cout<<s[i];
	//cout<<x<<" "<<tmp<<endl;
	if(tmp<=1ll*n*(n+1)/4) return 1;
	else return 0;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&a[i]);
		b[i]=a[i];
	}
	sort(b+1,b+n+1);
	int l=1,r=n+1;
	while(l<r-1)
	{
		int mid=(l+r)/2;
		//cout<<b[mid]<<" "<<endl;
		if(check(b[mid]))l=mid;
		else r=mid;
	}
	printf("%d",b[l]);
	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<bits/stdc++.h>
#define N 100005
using namespace std;
int n;
int a[N],b[N];
int s[N];
long long tmp;
int c[N];
void gb(int l,int r)
{
	if(l==r)return;
	int mid=(l+r)/2,h1=l,h2=mid+1,cnt=l;
	gb(l,mid);
	gb(mid+1,r);
	while(h1<=mid&&h2<=r)
	{
		if(s[h1]<=s[h2]) c[cnt++]=s[h1++];
		else c[cnt++]=s[h2++],tmp+=(mid-h1+1);
	}
	for(int i=h1;i<=mid;++i)c[cnt++]=s[i];
	for(int i=h2;i<=r;++i)c[cnt++]=s[i];
	for(int i=l;i<=r;++i) s[i]=c[i];
	//cout<<l<<" "<<r<<" "<<tmp<<endl;
}
bool check(int x)
{
	memset(s,0,sizeof(s));
	for(int i=1;i<=n;++i)
		if(a[i]<x) s[i]=s[i-1]-1;
		else s[i]=s[i-1]+1;
	//for(int i=0;i<=n;++i)
	//cout<<s[i];
	//	cout<<endl;
	tmp=0;gb(0,n);
	//for(int i=0;i<=n;++i)cout<<s[i];
	//cout<<x<<" "<<tmp<<endl;
	if(tmp<=1ll*n*(n+1)/4) return 1;
	else return 0;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&a[i]);
		b[i]=a[i];
	}
	sort(b+1,b+n+1);
	int l=1,r=n+1;
	while(l<r-1)
	{
		int mid=(l+r)/2;
		//cout<<b[mid]<<" "<<endl;
		if(check(b[mid]))l=mid;
		else r=mid;
	}
	printf("%d",b[l]);
	return 0;
}
```