#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;i++)
#define fdr(i,a,b) for(int i=a;i>=b;i--)
#define xx 100001
#define ll long long
inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
	for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}
ll a[xx],cha[xx],ru[xx*2];
ll n;
inline void cot(ll k,ll x)
{
	while(k<xx*2)
	{
		ru[k]+=x;
		k+=(k&-k);
	}
}
inline ll get(ll k)
{
	ll ans=0;
	while(k>0)
	{
		ans+=ru[k];
		k-=(k&-k);
	}
	return ans;
}
inline bool check(ll k)
{
	memset(ru,0,sizeof(ru));
	cha[0]=0;
	fur(i,1,n) cha[i]=cha[i-1]+(a[i]>=k?1:-1);
	ll ans=0;
	fur(i,0,n)
	{
		ans+=get(cha[i]+xx);
		cot(cha[i]+xx,1ll);
	}
	return ans>=(n*(n+1)>>2);
}
int main()
{
	ll l=0,r=0;
	n=in;
	fur(i,1,n) a[i]=in,r=max(a[i],r);
	while(l<=r)
	{
		ll mid=l+r>>1;
		if(check(mid)) l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",r);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define in read()
#define fur(i,a,b) for(int i=a;i<=b;i++)
#define fdr(i,a,b) for(int i=a;i>=b;i--)
#define xx 100001
#define ll long long
inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	for(;!isalnum(ch);ch=getchar()) if(ch=='-') f=-1;
	for(;isalnum(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}
ll a[xx],cha[xx],ru[xx*2];
ll n;
inline void cot(ll k,ll x)
{
	while(k<xx*2)
	{
		ru[k]+=x;
		k+=(k&-k);
	}
}
inline ll get(ll k)
{
	ll ans=0;
	while(k>0)
	{
		ans+=ru[k];
		k-=(k&-k);
	}
	return ans;
}
inline bool check(ll k)
{
	memset(ru,0,sizeof(ru));
	cha[0]=0;
	fur(i,1,n) cha[i]=cha[i-1]+(a[i]>=k?1:-1);
	ll ans=0;
	fur(i,0,n)
	{
		ans+=get(cha[i]+xx);
		cot(cha[i]+xx,1ll);
	}
	return ans>=(n*(n+1)>>2);
}
int main()
{
	ll l=0,r=0;
	n=in;
	fur(i,1,n) a[i]=in,r=max(a[i],r);
	while(l<=r)
	{
		ll mid=l+r>>1;
		if(check(mid)) l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",r);
	return 0;
}
```