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>
//CLOCKS_PER_SEC
#define se second
#define fi first
#define ll long long
#define Pii pair<int,int>
#define Pli pair<ll,int>
#define ull unsigned long long
#define pb push_back
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
const int N=1e5+10;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
using namespace std;
int n,val[N<<1];
int a[N],b[N];
int sum[N];
void add(int x){
	while(x<N*2){
		val[x]+=1;
		x+=x&-x;
	}
}
ll get(int x){
	ll ans=0;
	while(x>0){
		ans+=val[x];x-=x&-x;
	}
	return ans;
}
bool check(int x){
	for(int i=1;i<=n;++i)sum[i]=sum[i-1]+(a[i]>=x?1:-1);
	ll tot=0;
	memset(val,0,sizeof(val));
	for(int i=0;i<=n;++i){
		tot+=get(sum[i]+N);
		add(sum[i]+N);
	}
	return tot>=1LL*n*(n+1)/4;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",&a[i]);b[i]=a[i];
	}
	sort(b+1,b+1+n);
	int l=1,r=n;
	int ans=1;
	while(l<=r){
		int mid=(l+r)>>1;
		if(check(b[mid])){
			ans=mid;
			l=mid+1;
		}
		else r=mid-1;
	}
	cout<<b[ans]<<endl;
	return 0;
}  