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)

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <stdio.h>
#include <string.h>

using namespace std;

static const double EPS = 1e-9;
template<class T> bool INRANGE(T x, T a, T b) { return a <= x && x <= b; }
template<class T> void amin(T &a, T v) { if (a > v) a = v; }
template<class T> void amax(T &a, T v) { if (a < v) a = v; }
int ROUND(double x) { return (int)(x + 0.5); }
bool ISINT(double x) { return fabs(ROUND(x) - x) <= EPS; }
bool ISEQUAL(double x, double y) { return fabs(x - y) <= EPS * max(1.0, max(fabs(x), fabs(y))); }
double SQSUM(double x, double y) { return x * x + y * y; }
#define PI  (acos(-1))
#define ARRAY_NUM(a) (sizeof(a)/sizeof(a[0])) 
#define NG (-1)
#define BIG ((int)1e9+10)
#define BIGLL ((ll)4e18)
#define SZ(a) ((int)(a).size())
#define SQ(a) ((a)*(a))
typedef unsigned long long ull;
typedef long long ll;


#define MAXN 100005

int n, a[MAXN], b[2 * MAXN], s;
ll total;

int query(int p) {
	int t = 0;
	for (int i = (p + MAXN); i; i -= (i & -i)) {
		t += b[i];
	}
	return t;
}
void update(int p) {
	for (int i = (p + MAXN); i < 2 * MAXN; i += (i & -i)) {
		b[i]++;
	}
}

ll getTotal(int x, const vector <int>& c)
{
	total = 0;
	s = 0;
	memset(a, 0, sizeof(a));
	memset(b, 0, sizeof(b));

	for (int i = 0; i<n; ++i) {
		a[i] = (c[i] >= x ? 1 : -1);
	}
	update(0);

	for (int i = 0; i<n; ++i) {
		s += a[i];
		total += query(s);
		update(s);
	}

	return total;
}

int main() 
{
	cin >> n;

	vector <int> c(n);

	for (int i = 0; i<n; ++i) {
		cin >> c[i];
	}


	vector <int> d(c);
	sort(d.begin(), d.end());

/*
	for (int i = 0; i < n; ++i)
	{
		ll aboveXTotal = getTotal(d[i], c);
		// BEGIN CUT HERE
		cerr << " i=" << i << " d[i]=" << d[i] << " aboveXTotal=" << aboveXTotal << endl;
		// END CUT HERE


	}
	*/

	ll rank = (ll)n * (n + 1LL) / 2 / 2;

	ll lo = 0;
	ll hi = n - 1;
	// BEGIN CUT HERE
//	cerr << " rank=" << rank << endl;
	// END CUT HERE

	while (lo<hi)
	{
		ll mid = lo + (hi - lo + 1) / 2LL; // +1 is necessary, otherwise infinite loop when only two elements.

		ll aboveXTotal = getTotal(d[mid], c);


		// BEGIN CUT HERE
//		cerr << " lo=" << lo << " hi=" << hi << " mid=" << mid << " d[mid]=" << d[mid] << " aboveXTotal=" << aboveXTotal << endl;
		// END CUT HERE




		if (aboveXTotal < rank)
		{
			hi = mid - 1;
		}
		else
		{
			lo = mid;
		}
	}


	cout << d[lo] << endl;



	return 0;
}