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<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100010
#define lowbit(i) i&-i
using namespace std;
typedef long long ll;
int n;
int l=0,r=0;
int a[maxn],pre[maxn*10],t[maxn*10];
int readint(){
	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*10+ch-'0';
		ch=getchar();
	}
	return x*f;
}
void add(int i){
	while(i<=maxn*2){
		t[i]++;
		i+=lowbit(i);
	}
}
ll query(int i){
	ll ret=0;
	while(i>0){
		ret+=t[i];
		i-=lowbit(i);
	}
	return ret;
}
bool check(int mid){
	for(int i=1;i<=2*maxn;i++) t[i]=0;
	pre[0]=0;
	for(int i=1;i<=n;i++) pre[i]=pre[i-1]+(a[i]>=mid?1:-1);
	ll sum=0;
	for(int i=0;i<=n;i++){
		sum+=query(pre[i]+maxn);
		add(pre[i]+maxn);
	}
	return sum>=1ll*n*(n+1)/4;
}
int main(){
	n=readint();
	for(int i=1;i<=n;i++){
		a[i]=readint();
		r=max(r,a[i]);
	}
	int mid;
	while(l<=r){
		mid=(l+r)>>1;
		if(check(mid))
			l=mid+1;
	    else
			r=mid-1;
	}
	printf("%d",r);
	return 0;
}
