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<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define int long long
#define MN 100005
using namespace std;
inline int in(){
    int x=0;bool f=0;char c;
    for (;(c=getchar())<'0'||c>'9';f=c=='-');
    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');
    return f?-x:x;
}
int c[MN<<2],d[MN],sum[MN],a[MN],b[MN];
int n,tot,l,r,res;
ll num;
inline void add(int x,int v){
	for (int i=x;i<=((n<<1)|1);i+=i&(-i)) c[i]+=v; 
}
inline int query(int x){
	int res=0;
	for (int i=x;i;i-=i&(-i)) res+=c[i];
	return res;
}
inline bool check(int x){
	for (int i=1;i<=n;++i) sum[i]=sum[i-1]+(a[i]>b[x]);
	for (int i=1;i<=n;++i) d[i]=(sum[i]<<1)-i+n+1;d[0]=n+1;
	memset(c,0,sizeof(c));
	add(d[0],1);num=0ll;
	for (int i=1;i<=n;++i)
	num+=1ll*(query(((n<<1)|1))-query(d[i])),add(d[i],1);
	return (num>1ll*(n*(n+1ll))/4ll);
}
signed main()
{
	n=in();
	for (int i=1;i<=n;++i) a[i]=b[i]=in();
	sort(b+1,b+n+1);
	tot=unique(b+1,b+n+1)-b-1;
	l=0,r=tot,res=0;
	while (l<=r){
		int mid=(l+r)>>1;
		if (check(mid)) res=mid,r=mid-1;
		else l=mid+1;
	}
	printf("%d",b[res]);return 0;
}