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)

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define trv(p, u) for (edg *p = h[u]; p; p = p->nxt)
#define push_back pb
typedef long long ll;
typedef double db;
const int N = 2e5 + 100;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

inline int rd() {
  int s = 0, w = 1; char c = getchar();
  while (c < '0' || c > '9') { if (c == '-') w = -1; c = getchar(); }
  while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
  return s * w;
}

int n, ans, a[N], s[N];

struct nod {
  int lft, rgt, sum;
  nod *ls, *rs;
  inline void setc() { sum++; }
  inline void upd() { sum = ls->sum + rs->sum; }
} pol[N << 3], *rot, *cur = pol;

inline void build(nod *r, int lft, int rgt) {
  r->lft = lft, r->rgt = rgt;
  if (lft == rgt) {
    r->sum = 0;
    return ;
  }
  int mid = (lft + rgt) >> 1;
  nod *ls = ++cur, *rs = ++cur;
  r->ls = ls, r->rs = rs;
  build(ls, lft, mid), build(rs, mid + 1, rgt);
  r->upd();
}

inline void chg(nod *r, int pos) {
  if (r->lft == pos && r->rgt == pos) {
    r->setc(); return ;
  }
  if (r->ls->rgt >= pos) chg(r->ls, pos);
  else chg(r->rs, pos);
  r->upd();
}

inline int qry(nod *r, int lft, int rgt) {
  if (lft > rgt) return 0;
  if (r->lft == lft && r->rgt == rgt) return r->sum;
  if (r->ls->rgt >= rgt) return qry(r->ls, lft, rgt);
  else if (r->rs->lft <= lft) return qry(r->rs, lft, rgt);
  else return qry(r->ls, lft, r->ls->rgt) +
              qry(r->rs, r->rs->lft, rgt);
}

inline bool chk(int x) {
  ll res = 0;
  memset(s, 0, sizeof(s));
  cur = pol;
  rep(i, 1, n) {
    if (a[i] <= x) s[i] = s[i - 1] + 1;
    else s[i] = s[i - 1] - 1;
  }
  rep(i, 0, n) s[i] += (n + 1);
  build(rot = pol, 1, N - 100);
  rep(i, 0, n) {
    res += 1ll * qry(rot, 1, s[i] - 1);
    chg(rot, s[i]);
  }
  return (res >= 1ll * n * (n + 1) / 4 + 1);
}

int main() {
  n = rd();
  rep(i, 1, n) a[i] = rd();
  int l = 1, r = inf;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (chk(mid)) ans = mid, r = mid - 1;
    else l = mid + 1;
  }
  printf("%d\n", ans);
  return 0;
}