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
import java.util.Arrays;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int N = scanner.nextInt();
    int[] a = new int[N];
    for (int i = 0; i < N; i++) {
      a[i] = scanner.nextInt();
    }
    int[] sorted = Arrays.copyOf(a, a.length);
    Arrays.sort(sorted);

    long target = N;
    target = (target * (N + 1) / 2 + 1) / 2;
    int left = -1, right = N;
    while (right - left > 1) {
      int mid = (left + right) / 2;
      long count = numSubarraysWhooseMedianBiggerThan(a, sorted[mid]);
      if (target <= count) {
        left = mid;
      } else {
        right = mid;
      }
    }
    System.out.println(sorted[left]);
  }

  private static long numSubarraysWhooseMedianBiggerThan(int[] a, int t) {
    int[] b = new int[a.length + 1];
    b[0] = 0;
    int sum = 0;
    for (int i = 0; i < a.length; i++) {
      sum += a[i] >= t ? 1 : -1;
      b[i + 1] = sum;
    }
    long n = a.length;
    return n * (n + 1) / 2 - countInversion(b, 0, b.length);
  }

  private static long countInversion(int[] a, int left, int right) {
    if (left + 1 >= right) return 0;
    int mid = (left + right) / 2;
    long count = countInversion(a, left, mid) + countInversion(a, mid, right);
    int[] b = new int[right - left];
    int i = 0;
    int li = left;
    int ri = mid;
    while (li < mid && ri < right) {
      if (a[li] <= a[ri]) {
        b[i++] = a[li++];
      } else {
        b[i++] = a[ri++];
        count += mid - li;
      }
    }
    while (li < mid) b[i++] = a[li++];
    while (ri < right) b[i++] = a[ri++];
    System.arraycopy(b, 0, a, left, b.length);
    return count;
  }
}
