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.

Examples

Input

3
10 30 20


Output

30


Input

1
10


Output

10


Input

10
5 9 5 9 8 9 3 5 4 3


Output

8
#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define all(a) (a).begin(), (a).end()

using namespace std;
using i32 = int;
using vi32 = vector<i32>;
using i64 = long long;
using vi64 = vector<i64>;

i64 n;
vi32 a;
vi64 c;

void add(int x) {
  while (x <= 2 * n) {
    c[x]++;
    x += x & -x;
  }
}

i64 sum(int x) {
  i64 s = 0;
  while (x) {
    s += c[x];
    x -= x & -x;
  }
  return s;
}

bool check(int md) {
  vi64 s(n + 1);
  c = vi64(n * 3);
  rep(i, n) {
    if (a[i] >= md)
      s[i + 1] = s[i] + 1;
    else
      s[i + 1] = s[i] - 1;
  }
  i64 cnt = 0;
  rep(i, n + 1) {
    cnt += sum(s[i] + n);
    add(s[i] + n);
  }
  return cnt >= n * (n + 1) / 4;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(16);
  cerr << fixed << setprecision(16);
  cin >> n;
  a = vi32(n);
  rep(i, n) cin >> a[i];
  int ok = *min_element(all(a)), ng = *max_element(all(a)) + 1;
  while (ng - ok > 1) {
    int md = (ok + ng) / 2;
    if (check(md)) {
      ok = md;
    } else {
      ng = md;
    }
  }
  cout << ok << endl;
  return 0;
}
