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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
const int maxn=1e5+10;
using namespace std;
typedef long long ll;
ll n,a[maxn],tree[maxn*2+5],pre[maxn*2+5],l,r;
void add(ll x,ll v)
{
	for(ll i=x;i<=2*maxn;i+=i&-i)
    	tree[i]+=v;
}
ll ask(int x)
{
    ll ans=0;
    for(ll i=x;i>0;i-=i&-i)
    	ans+=tree[i];
    return ans;
}
bool check(ll mid)
{	
	ll ans=0;
	memset(tree,0,sizeof(tree));
	pre[0]=0;
	for(ll i=1;i<=n;++i)
	pre[i]=pre[i-1]+(a[i]>=mid?1:-1);
	for(ll i=0;i<=n;++i)
	{
		ans+=ask(pre[i]+maxn);
		add(pre[i]+maxn,1ll);
	}
	return ans>=n*(n+1)/4;
}
int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%lld",&a[i]);
	    r=max(r,a[i]);
    }
    while(l<=r)
    {
    	ll mid=(l+r)>>1; 
    	if(check(mid))l=mid+1;
		else r=mid-1; 
    }
    printf("%lld",r);
	return 0;
}