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>
using namespace std;

struct BIT {
  long long n;
  vector<long long> dat;

  void init(long long n_input) {
    n = n_input + 1;
    dat.resize(n + 1);
    for (int i = 0; i <= n; i++) dat[i] = 0;
  }

  long long sum(long long i) {
    long long s = 0;
    while (i >= 0) {
      s += dat[i];
      i = (i & (i + 1)) - 1;
    }
    return s;
  }

  void add(long long i, long long x) {
    while (i <= n) {
      dat[i] += x;
      i |= i + 1;
    }
  }
};

int main() {
  int i;
  long long N;
  cin >> N;
  vector<int> A(N), sortedA(N);
  for (i = 0; i < N; i++) {
    cin >> A[i];
    sortedA[i] = A[i];
  }
  sort(sortedA.begin(), sortedA.end());
  int ub = N;
  int lb = 0;
  BIT bit;

  long long border = ((N * (N + 1) / 2) + 1) / 2;
  while (ub - lb > 1) {
    int mid = (ub + lb) / 2;
    int midn = sortedA[mid];
    bit.init(2 * N + 1);
    int sum = 0;
    bit.add(sum + N, 1);
    long long result = 0;
    for (i = 0; i < N; i++) {
      sum += A[i] >= midn ? 1 : -1;
      result += bit.sum(sum + N);
      bit.add(sum + N, 1);
    }
    if (result >= border) {
      lb = mid;
    } else {
      ub = mid;
    }
  }
  cout << sortedA[lb] << endl;
  return 0;
}
