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 ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define swap(x,y) (x^=y^=x^=y)
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 100005 
int n,m,a[MN],b[MN],c[MN];
int t[MN<<1];
void C(int x){for(;x<MN*2;x+=(x&-x))t[x]++;}
int G(int x){int res=0;for(;x;x-=(x&-x))res+=t[x];return res;}
inline bool check(int x)
{
	ll cnt=0ll;
	register int i;
	for(i=1;i<=n;++i) c[i]=c[i-1]+(a[i]>=x?1:-1);
	memset(t,0,sizeof t);
	for(i=1;i<=n;++i)
	{
		C(c[i-1]+MN);
		cnt+=G(c[i]+MN);
	}
	return cnt*4>=1ll*n*(n+1);
}
int main()
{
	n=read();register int i;
	for(i=1;i<=n;++i) a[i]=b[i]=read();
	std::sort(b+1,b+n+1);
	m=std::unique(b+1,b+n+1)-b-1;
	int l,r,ans;
	for(l=1,r=m,ans=0;l<=r;)
	{
		int mid=l+r>>1;
		if(check(b[mid])) ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",b[ans]);
}