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<algorithm>
#include<cstring>
#define ll long long
using namespace std;
ll input[200005];
ll n;
ll  l,r,ans,mid;
ll  lib[200005];
ll sum[200005];
ll  tree[200005];
inline ll lowbit(ll x) {
	return x&(-x);
	}
ll minn=0;
ll maxx=0;
void add(ll x) {
	for(ll i=x; i<=maxx; i+=lowbit(i))tree[i]+=1;
	}
ll summ(ll x) {
	ll kk=0;
	for(ll i=x; i; i-=lowbit(i))kk+=tree[i];
	return kk;
	}
bool check(ll x) {
	memset(tree,0,sizeof(tree));
	ll anss=0;
	minn=0;
	maxx=0;
	for(ll i=1; i<=n; i++) {
		if(input[i]>=x)sum[i]=1;
		else sum[i]=-1;
		}
	for(ll i=1; i<=n; i++)
		sum[i]+=sum[i-1],minn=min(minn,sum[i]);
	for(ll i=0; i<=n; i++)
		sum[i]=sum[i]-minn+1,maxx=max(maxx,sum[i]);
	for(ll i=0; i<=n; i++) {
		anss+=summ(sum[i]);
		add(sum[i]);
		}
	return anss>=n*(n+1)/4;
	}
int main() {
	scanf("%lld",&n);
	r=0;
	for(ll i=1; i<=n; i++) {
		scanf("%lld",&input[i]);
		r=max(r,input[i]);
		}
	l=1;
	while(l<=r) {
		mid=(l+r)>>1;
		if(check(mid)) {
			ans=mid;
			l=mid+1;
			}
		else r=mid-1;
		}
	cout<<ans;
	}
