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: Runtime Error)

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define INF 10000000
#define MOD 1000000007
#define MID (l+r)/2
#define HASHMOD 2305843009213693951
#define ll long long
#define ull unsigned long long
#define F first
#define S second
typedef pair<ll, ll> ii;
typedef pair<ii, ll> iii;
typedef vector<ll> vi;
typedef vector<ii> vii;
typedef map<ll, ll> mii;

#define EPS 1e-6

int b[10*INF];
int seg[10*INF];

ll query(int rl, int rr, int l=0, int r=INF-1, int ind=1){
	if(rl<=l && r<=rr) return seg[ind];
	else if(rl>r || l>rr) return 0;
	return query(rl, rr, l, MID, ind*2) + query(rl,rr,MID+1,r,ind*2+1);
}

void update(int x, int l=0, int r=INF-1, int ind=1){
	if(l==r && l==x){ seg[ind]++; return;}
	if(x<l || x>r) return;
	update(x,l,MID,ind*2);
	update(x,MID+1,r,ind*2+1);
	seg[ind] = seg[ind*2] + seg[ind*2+1];
}


int main(){
	int n;
	cin>>n;
	ll a[n];
	for(int i=0;i<n;i++) cin>>a[i];
	ll l=0, r=2000000000;
	ll m;
	while(l<=r){
		m = MID;
		//cout<<m<<endl;
		int ans=0;
		for(int i=0;i<n;i++){
			b[i] = (a[i]>m ? 1 : (a[i]==m ? 0 : -1));
			if(b[i]>=0) ans++;
		}
		for(int i=1;i<n;i++) b[i]+=b[i-1];
		int mina = b[0];
		for(int i=0;i<n;i++) mina = min(mina, b[i]);
		//cout<<mina<<endl;
		for(int i=0;i<n;i++) b[i]-=mina-1;
		for(int i=0;i<n;i++) assert(b[i]<1000000);
		memset(seg, 0, 9*INF);
		//cout<<endl;
		for(int i=0;i<n;i++){
			//cout<<query(0,b[i])<<" ";
			ans+=query(0, b[i]);
			update(b[i]);
		}
		//cout<<endl;
		//cout<<ans<<endl;
		//cout<<endl;
		if(ans <= (n*(n-1))/4) r = m-1;
		else l = m+1;
	}
	cout<<m<<endl;
}