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>
using namespace std;
const int maxn=200003,D=100000,INF=1050000000;
int t[maxn],n,a[maxn],sum[maxn];
void add(int pos){while(pos<maxn)t[pos]++,pos+=pos&-pos;}
int query(int pos){int ret=0;while(pos)ret+=t[pos],pos-=pos&-pos;return ret;}
bool check(int x){
	for(int i=1;i<=n;i++)sum[i]=sum[i-1]+(a[i]>=x?1:-1);
	long long tot=0;
	memset(t,0,sizeof(t));
// printf("chk %d\n",x);
	for(int i=1;i<=n;i++){
		add(sum[i-1]+D);
		tot+=query(sum[i]+D);
// printf("query %d: %d\n",sum[i],query(sum[i]+D));
		if(tot>=(long long)n*(n+1)/4)return 1;
	}
	return 0;
}
int main(){
	int l=INF,r=0,mid,ans;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",a+i);
		l=min(l,a[i]);
		r=max(r,a[i]);
	}
	while(l<=r){
		mid=(l+r)>>1;
		if(check(mid))ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",ans);
	return 0;
}