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)

#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;

/*BigInteger
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/rational.hpp>
namespace xxx = boost::multiprecision;
using Bint = xxx::cpp_int;
using Real = xxx::number<xxx::cpp_dec_float<1024>>;
*/

#define int long long
#define pb(x) push_back(x)
#define m0(x) memset((x), 0LL, sizeof(x))
#define mm(x) memset((x), -1LL, sizeof(x))

//container
#define ALL(x) (x).begin(), (x).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define UNIQUE(v) (v).erase(unique((v).begin(), (v).end()), (v).end());
#define PERM(c) \
  sort(ALL(c)); \
  for (bool c##p = 1; c##p; c##p = next_permutation(ALL(c)))

// debug
#define GET_VAR_NAME(variable) #variable
#define test(x) cout << GET_VAR_NAME(x) << " = " << x << endl;

// bit_macro
#define Bit(n) (1LL << (n))
#define Bitset(a, b) (a) |= (1LL << (b))
#define Bitunset(a, b) (a) &= ~(1LL << (b))
#define Bitcheck(a, b) ((((a) >> (b)) & 1LL) == 1LL)
#define Bitcount(a) __builtin_popcountll((a))

//typedef
typedef long long lint;
typedef unsigned long long ull;
typedef complex<long double> Complex;
typedef pair<int, int> P;
typedef tuple<int, int, int> TP;
typedef vector<int> vec;
typedef vector<vec> mat;

//constant
constexpr int INF = (int)1e18;
constexpr int MOD = (int)1e9 + 7;
constexpr double PI = (double)acos(-1);
constexpr double EPS = (double)1e-10;
constexpr int dx[] = {-1, 0, 0, 1, 0, -1, -1, 1, 1};
constexpr int dy[] = {0, -1, 1, 0, 0, 1, -1, 1, -1};

//
template <typename T>
void chmax(T &a, T b) { a = max(a, b); }
template <typename T>
void chmin(T &a, T b) { a = min(a, b); }
//
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}

//
struct Accelerate_Cin {
  Accelerate_Cin() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(20);
  };
} Accelerate_Cin;

int N;
vector<int> a;
int n;

int bit[200001];
int sum(int i) {
  int res = 0;
  while (i > 0) {
    res += bit[i];
    i -= i & -i;
  }
  return res;
}

void add(int i, int x) {
  while (i <= n) {
    bit[i] += x;
    i += i & -i;
  }
}

bool C(int m) {
  m0(bit);

  vector<int> aa(N);
  for (int i = 0; i < N; i++) {
    if (a[i] < m) {
      aa[i] = -1;
    } else if (a[i] >= m) {
      aa[i] = 1;
    }
  }

  vector<int> acc(N + 1, 0);
  acc[0] = 100000;
  for (int i = 0; i < N; i++) acc[i + 1] = aa[i] + acc[i];

  int ans = 0;
  for (int i = 0; i < N + 1; i++) {
    ans += sum(acc[i]);
    add(acc[i], 1);
  }

  if (2 * ans >= N * (N + 1) / 2) {
    return true;
  } else {
    return false;
  }
}

signed main() {
  cin >> N;
  a.resize(N);
  n = 202020;
  vector<int> A(N);
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    A[i] = a[i];
  }

  sort(ALL(A));

  int ng = N, ok = -1;
  while (abs(ok - ng) > 1) {
    int mid = (ok + ng) / 2;
    if (C(A[mid])) {
      ok = mid;
    } else {
      ng = mid;
    }
  }

  cout << A[ok] << endl;
  return 0;
}