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 <iostream>
#include <vector>
#include <algorithm>
#define int long long
using namespace std;

signed main() {
  int N; cin >> N;
  vector<int> a(N), s(N), l(N+1);
  l[0] = 0;
  for(int i = 0; i < N; ++i) {
    cin >> a[i];
    s[i] = a[i];
  }
  sort(s.begin(), s.end());
  int right = N-1;
  int left = 0;
  while(right != left) {
    int j = (right + left + 1) / 2;
    int num = s[j];
    for(int i = 0; i < N; ++i) {
      if(a[i] >= num) {
        l[i+1] = 1;
      } else {
        l[i+1] = -1;
      }
    }
    for(int i = 1; i <= N; ++i) l[i] += l[i-1];


    int count, t = 0;
    vector<int> bit(2*N+2, 0);
    for(int i = 0; i < N + 1; ++i) {
      int first = l[i] + N + 1;
      for(int x = first; x <= 2 * N + 1; x += x&-x) {
        bit[x]++;
      }
      int sum = 0;
      for(int x = first; x >= 1; x-=x&-x) sum += bit[x];
      t += i+1-sum;
    }
    count = N * (N + 1) / 2 - t;
    if(count >= (N * (N+1)/2+1)/2) {
      left = j;
    } else {
      right = j - 1;
    }
  }
  cout << s[left] << endl;

  return 0;
}
