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: Wrong Answer)

#pragma GCC optimize(2)
// Happy TLE and WA every day!
// by: zxb the vegetable chicken
#include<bits/stdc++.h>
#define mp				make_pair
#define rep(i,n)		for(int i = 0; i < n; i++)
#define foreach(i,c)	for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define BINF			0x7fffffff
#define INF				0x3f3f3f3f
#define LINF			3223372036854775807
#define END(s)			{cout<<s; return 0;}
#define CON(s)			{cout<<s; continue;}
#define BRE(s)			{cout<<s; break;}
#define pb				push_back
//#define int				long long
#define All(a)			a.begin(), a.end()  //with A in CAPITAL!!!
#define sz(a)			(int)a.size()
#define F				first
#define S				second
//#define usingFiles
using namespace std;

const int rp = 666666;
const bool debug = 1;
const bool I_good_vegetable_a = 1;
const int S = 111111;

int bit[444444], n;
int in[100100];
int a[100100];
int s[100100];
map<int, bool> has;

int sum(int now){
	int Sum = 0;
	while(now > 0){
		Sum += bit[now];
		now -= (now & -now);
	}
	return Sum;
}

void change(int now, int x){
	while(now <= n + 2 * S){
		bit[now] += x;
		now += (now & -now);
	}
}

int SRA(){
	memset(bit, 0, sizeof(bit));
	int ans = 0;
	rep(j, n+1){
		ans += j - sum(s[j]);
		change(s[j], 1);
	}
	return ans;
}

// ±ÈnowÐ¡µÄÖÐÎ»Êý¸öÊý 
int check(int now){
	s[0] = S;
	rep(i, n){
		if(in[i] > now) a[i] = 1;
		else if(in[i] < now) a[i] = -1;
		else a[i] = 0;
		s[i+1] = a[i] + s[i];
	}
	return SRA();
}

signed main(){
	ios::sync_with_stdio(false);
	#ifdef usingFiles
		freopen("input.txt", "r", stdin);
		freopen("output.txt", "w", stdout);
	#endif
	cin >> n;
	rep(i, n){
		cin >> in[i];
		has[in[i]] = true;
	}
	int num = n * (n + 1) / 2;
	int ans = 0;
	int l = -1, r = 1e9 + 7;
	while(l <= r){
		int mid = (l+r) / 2;
		int asdf = check(mid);
		if(asdf > num/2) r = mid-1;
		else if(asdf == num/2){
			ans = mid;
			break;
		}
		else l = mid+1;
	}
	if(check(l) == num/2) ans = l;
	else if(check(r) == num/2) ans = r;
	else ans = (l + r) / 2;
	while(!has.count(ans)) ans++;
	if(check(ans) != num/2){
		ans--;
		while(!has.count(ans)) ans--;
	}
	cout << ans << endl;
	return 0;
}