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.io.BufferedInputStream;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(new BufferedInputStream(System.in));
        int N = scan.nextInt();
        int[] arr = new int[N];
        for (int i=0;i<N;i++) arr[i] = scan.nextInt();
        Task tsk = new Task();
        System.out.println(tsk.solve(arr));
    }

    static class Task{
        public int solve(int[] arr){
            int n = arr.length;
            int[] sorted = Arrays.copyOf(arr, n);
            Arrays.sort(sorted);

            long total = (long)n * (n+1) / 2;
            int l = 0, r = n-1;
            while (l < r) {
                int mid = l + (r-l)/2;
                int target = sorted[mid];

                int cur = 0;
                long cnt = 0;
                long[] bit = new long[2*n+2];
                for (int x=n+1;x<bit.length;x+=x&-x) bit[x]++;
                for (int i=0;i<n;i++) {
                    cur += arr[i] <= target? 1: -1;
                    for (int x=cur+n+1;x<bit.length;x+=x&-x) bit[x]++;
                    for (int x=cur+n;x>0;x-=x&-x) cnt += bit[x];
                }
                if (cnt >= total/2 + 1) r = mid;
                else l = mid+1;
            }
            return sorted[l];
        }
    }
}
