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

struct Ftree {
  vector<int> dat;
  Ftree(int n) : dat(n + 1) {}
  void add(int x, int v) {
    for (int i = x + 1; i < dat.size(); i += i & -i) {
      dat[i] += v;
    }
  }
  int sum(int x) {
    if (x < 0) return 0;
    int res = 0;
    for (int i = x + 1; i; i -= i & -i) {
      res += dat[i];
    }
    return res;
  }
};

signed main() {
  ios::sync_with_stdio(false);

  int N;
  cin >> N;

  vector<int> A(N);
  for (int i = 0; i < N; ++i) {
    cin >> A[i];
  }

  int maxa = *max_element(A.begin(), A.end());
  int ans = -1; {
    int lb = 0;
    int ub = maxa + 1;
    while (ub - lb > 1) {
      int mb = lb + ub >> 1;
      vector<int> a = A;
      for (int &v : a) v = v < mb ? -1 : 1;
      vector<int> sa = vector<int>(a.size() + 1);
      for (int i = 0; i < a.size(); ++i) {
        sa[i + 1] = sa[i] + a[i];
      }
      Ftree tree(2 * N + 1);
      int64_t cnt = 0;
      for (int v : sa) {
        cnt += tree.sum(2 * N) - tree.sum(N + v);
        tree.add(N + v, 1);
      }
      (1LL*N*(N+1)/4 < cnt ? ub : lb) = mb;
    }
    ans = lb;
  }

  cout << ans << endl;

  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

struct Ftree {
  vector<int> dat;
  Ftree(int n) : dat(n + 1) {}
  void add(int x, int v) {
    for (int i = x + 1; i < dat.size(); i += i & -i) {
      dat[i] += v;
    }
  }
  int sum(int x) {
    if (x < 0) return 0;
    int res = 0;
    for (int i = x + 1; i; i -= i & -i) {
      res += dat[i];
    }
    return res;
  }
};

signed main() {
  ios::sync_with_stdio(false);

  int N;
  cin >> N;

  vector<int> A(N);
  for (int i = 0; i < N; ++i) {
    cin >> A[i];
  }

  int maxa = *max_element(A.begin(), A.end());
  int ans = -1; {
    int lb = 0;
    int ub = maxa + 1;
    while (ub - lb > 1) {
      int mb = lb + ub >> 1;
      vector<int> a = A;
      for (int &v : a) v = v < mb ? -1 : 1;
      vector<int> sa = vector<int>(a.size() + 1);
      for (int i = 0; i < a.size(); ++i) {
        sa[i + 1] = sa[i] + a[i];
      }
      Ftree tree(2 * N + 1);
      int64_t cnt = 0;
      for (int v : sa) {
        cnt += tree.sum(2 * N) - tree.sum(N + v);
        tree.add(N + v, 1);
      }
      (1LL*N*(N+1)/4 < cnt ? ub : lb) = mb;
    }
    ans = lb;
  }

  cout << ans << endl;

  return 0;
}

```