Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
10 30 20

Sample Output 1

30

The median of each contiguous subsequence of a is as follows:

The median of (10) is 10.

The median of (30) is 30.

The median of (20) is 20.

The median of (10, 30) is 30.

The median of (30, 20) is 30.

The median of (10, 30, 20) is 20.

Thus, m = (10, 30, 20, 30, 30, 20) and the median of m is 30.

Sample Input 2

1
10

Sample Output 2

10

Sample Input 3

10
5 9 5 9 8 9 3 5 4 3

Sample Output 3

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())

int n;
ll a[100005],rui[100005];

struct RMQ{
	#define s (1<<19)
	int seg[s];
	void init(){ memset(seg,0,sizeof(seg)); }
	void update(int k,int a){
		k+=s/2-1; seg[k]=a;
		while(k>0){
			k=(k-1)/2;
			seg[k]=(seg[k*2+1]+seg[k*2+2]);
		}
	}
	void update_add(int k,int a){
		k+=s/2-1; seg[k]+=a;
		while(k>0){
			k=(k-1)/2;
			seg[k]=(seg[k*2+1]+seg[k*2+2]);
		}
	}
	int query(int a,int b,int k,int l,int r){
		if(r<a || b<l) return 0;
		if(a<=l && r<=b) return seg[k];
		else{
			int vl=query(a,b,k*2+1,l,(l+r)/2);
			int vr=query(a,b,k*2+2,(l+r)/2+1,r);
			return (vl+vr);
		}
	}
}kaede;

int main(){
	cin >> n;
	repn(i,n) cin >> a[i];
	
	ll lb = 0, ub = INF+7;
	while(ub-lb > 1){
		ll mid = (lb+ub)/2;
		repn(i,n){
			rui[i] = rui[i-1];
			if(a[i] >= mid) rui[i] ++;
			else rui[i] --;
		}
		kaede.init();
		repn(i,n) kaede.update_add(rui[i]+100002,1);
		ll ans = 0;
		repn(i,n){
			//rui[i-1]以上?
			ans += kaede.query(rui[i-1]+100002,s/2-1,0,0,s/2-1);
			//rui[i]を消す
			kaede.update_add(rui[i]+100002,-1);
		}
		ll x = 1LL*n*(n+1)/2;
		x = (x+1)/2;
		if(x <= ans) lb = mid;
		else ub = mid;
	}
	
	cout << lb << endl;
}