#include <bits/stdc++.h>
#define int long long
using namespace std;

int n,ans,cnt,a[100010],s[100010],x[100010],y[100010];
map<int,int> mp;

inline void msort(int l,int r)
{
    if(l==r) return;
    int m=(l+r)>>1;
    msort(l,m);
    msort(m+1,r);
    
    int i=l,j=m+1;
    vector<int> v;
    while(i<=m&&j<=r)
    {
        if(a[i]<=a[j]) v.push_back(a[i++]),ans+=(j-m-1);
        else v.push_back(a[j++]);
    }
    
    while(i<=m) v.push_back(a[i++]),ans+=(j-m-1);
    while(j<=r) v.push_back(a[j++]);
    for(i=0;i<v.size();i++) a[l+i]=v[i];
}

bool check(int val){
	for(int i=1;i<=n;i++){
		if(s[i]<=val) y[i]=-1;
		else y[i]=1;
	}
	for(int i=1;i<=n;i++) a[i]=a[i-1]+y[i];
	ans=0;
	msort(1,n);
	int uu=0;
	for(int i=1;i<=n;i++){
		if(y[i]==-1) uu++;
		if(uu*2>i) ans++;
	}
	if(ans*2>n*(n+1)/2) return true;
	return false;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;i++){cin>>s[i];if(mp[s[i]]==0) x[++cnt]=s[i];mp[s[i]]=1;}
	sort(x+1,x+cnt+1);
	int l=1,r=cnt;
	while(l<=r)
	{
		int mid=(l+r)/2;
		if(check(x[mid])){
			r=mid-1;
		}
		else l=mid+1;
	}
	cout<<x[l]<<endl;
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 int long long
using namespace std;

int n,ans,cnt,a[100010],s[100010],x[100010],y[100010];
map<int,int> mp;

inline void msort(int l,int r)
{
    if(l==r) return;
    int m=(l+r)>>1;
    msort(l,m);
    msort(m+1,r);
    
    int i=l,j=m+1;
    vector<int> v;
    while(i<=m&&j<=r)
    {
        if(a[i]<=a[j]) v.push_back(a[i++]),ans+=(j-m-1);
        else v.push_back(a[j++]);
    }
    
    while(i<=m) v.push_back(a[i++]),ans+=(j-m-1);
    while(j<=r) v.push_back(a[j++]);
    for(i=0;i<v.size();i++) a[l+i]=v[i];
}

bool check(int val){
	for(int i=1;i<=n;i++){
		if(s[i]<=val) y[i]=-1;
		else y[i]=1;
	}
	for(int i=1;i<=n;i++) a[i]=a[i-1]+y[i];
	ans=0;
	msort(1,n);
	int uu=0;
	for(int i=1;i<=n;i++){
		if(y[i]==-1) uu++;
		if(uu*2>i) ans++;
	}
	if(ans*2>n*(n+1)/2) return true;
	return false;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;i++){cin>>s[i];if(mp[s[i]]==0) x[++cnt]=s[i];mp[s[i]]=1;}
	sort(x+1,x+cnt+1);
	int l=1,r=cnt;
	while(l<=r)
	{
		int mid=(l+r)/2;
		if(check(x[mid])){
			r=mid-1;
		}
		else l=mid+1;
	}
	cout<<x[l]<<endl;
	return 0;
}
```