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 rep(i,l,r) for(ll i=(l);i<=(r);++i)
#define per(i,r,l) for(ll i=(r);i>=(l);--i)
using namespace std;
const int maxn=200010;
typedef long long ll;
ll n,a[maxn],b[maxn],c[maxn],t,mn;
void add(ll i,ll x){for(;i<=t;i+=i&(-i))c[i]+=x;}
int sum(ll i){ll ans=0;for(;i;i-=i&(-i))ans+=c[i];return ans;}
bool check(ll x){
	ll mn=0,ans=0;memset(c,0,sizeof(c));
	rep(i,1,n) if(a[i]>=x) b[i]=1;else b[i]=-1;
	rep(i,1,n) b[i]+=b[i-1],mn=min(mn,b[i]);
	rep(i,0,n) b[i]+=-mn+1,t=max(t,b[i]);
	rep(i,0,n){
		ans+=sum(b[i]);
		add(b[i],1);
	}
	return ans>=1LL*n*(n+1)/4;
}
int main(){
	scanf("%d",&n);
	ll l=1,r=0;
	rep(i,1,n) scanf("%d",&a[i]),r=max(a[i],r);
	while(l<=r){
		ll mid=(l+r)>>1;
		if(check(mid)) l=mid+1;
		else r=mid-1;
	}
	printf("%d",r);
	return 0;
}