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
#include<bits/stdc++.h>
#define int long long

using namespace std;
typedef long long ll;
int n;

struct BIT
{
	ll c[(int)1e6+8];
	int lowbit(int x){return x&-x;}
	void add(int x,int val)
	{
		for(;x<=(int)1e6+5;x+=lowbit(x))
		{
			c[x]+=val;
		}
	}
	int query(int x)
	{
		int ret=0;
		for(;x>0;x-=lowbit(x))
		{
			ret+=c[x];
		}
		return ret;
	}
	void clear()
	{
		memset(c,0,sizeof(c));
	}
}T;

int a[(int)1e5+5];
int s[(int)1e5+5];
signed main()
{
	ios::sync_with_stdio(0);
	
	cin>>n;
	int ans=(n+1)*n/4+1;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	
	int l=0,r=1e9+1;
	while(l<r)
	{
		int m=l+r>>1;
		for(int i=1;i<=n;i++)
		{
			if(a[i]>m)s[i]=-1;else s[i]=1;
		}
		s[0]=0;
		for(int i=1;i<=n;i++)s[i]=s[i-1]+s[i];
		for(int i=0;i<=n;i++)s[i]+=1e5+5;
		T.clear();
		T.add(1e5+5,1);
		int cnt=0;
		for(int i=1;i<=n;i++)
		{
			cnt+=T.query(s[i]-1);
			T.add(s[i],1);
		}
		if(cnt>=ans)r=m;else l=m+1;
	}
	cout<<l;
	return 0;
}

//http://www.cnblogs.com/chy-2003/p/9751361.html