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
using namespace std;
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<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
int a[100005],n,b[100005],c[100005],bit[200005];
ll sum,res;
void add(int x){for(;x<=200002;x+=x&-x)bit[x]++;}
int qur(int x){int tmp=0;for(;x;x-=x&-x)tmp+=bit[x];return tmp;}
bool check(int k){
	res=0,sum=100002;memset(bit,0,sizeof bit);
	for(int i=1;i<=n;++i)if(a[i]<=k)b[i]=1;else b[i]=-1;
	add(sum);
	for(int i=1;i<=n;++i){
		sum+=b[i];
		res+=qur(sum-1);
		add(sum);
	}
	return res>=(1ll*n*(n+1)>>2)+1;
}
int main(){
	n=read();for(int i=1;i<=n;++i)a[i]=c[i]=read();
	sort(c+1,c+n+1);
	int l=1,r=n,mid;
	while(l<r){
		mid=l+r>>1;
		if(check(c[mid]))r=mid;
		else l=mid+1;
	}
	printf("%d",c[r]);
}