Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
10 30 20

Sample Output 1

30

The median of each contiguous subsequence of a is as follows:

The median of (10) is 10.

The median of (30) is 30.

The median of (20) is 20.

The median of (10, 30) is 30.

The median of (30, 20) is 30.

The median of (10, 30, 20) is 20.

Thus, m = (10, 30, 20, 30, 30, 20) and the median of m is 30.

Sample Input 2

1
10

Sample Output 2

10

Sample Input 3

10
5 9 5 9 8 9 3 5 4 3

Sample Output 3

8

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new PrintStream(System.out));
        n=Integer.parseInt(f.readLine());
        arr=new int[n];
        StringTokenizer st=new StringTokenizer(f.readLine());
        for(int i=0;i<n;i++){
            arr[i]=Integer.parseInt(st.nextToken());
        }
        long lo = 0;
        long hi = Integer.MAX_VALUE*5;
        while (hi > lo+1) {
            long mid = (lo + hi) / 2;
            if (doable(mid)) {
                lo = mid;
            } else {
                hi = mid;
            }
        }
        System.out.print(lo);
        f.close();
        out.close();
    }

    static int[]arr;
    static int n;
    static boolean doable(long cur){
        //nums such that greater
        BIT=new int[250000];
        int[]temp=new int[n];
        int[]pfx=new int[n];
        for(int i=0;i<n;i++){
            temp[i]=arr[i]>=cur?1:-1;
            pfx[i]=i==0?temp[i]:pfx[i-1]+temp[i];
//            update(i,pfx[i],120000);

        }
        long inv=0;
        //count inversions
        //more 1's than -1 if prefix condition is satisfied
        update(0,1,220000);
        for(int i=0;i<n;i++){
            inv+=query(pfx[i]);
            update(pfx[i],1,220000);
        }
        return inv>=(n*(long)(n-1)/4)+1;
    }
    static int[]BIT;
    //add val at index x
    //declare n very large
    static void update(int x, long val,int n)
    {
        x+=100005;
        for(; x <= n; x += x&-x)
            BIT[x] += val;
    }
    static long query(int x)
    {
        x+=100005;
        long sum = 0;
        for(; x > 0; x -= x&-x)
            sum += BIT[x];
        return sum;
    }
}
class pair implements Comparable <pair>{
    int num;
    int idx;

    public int compareTo(pair other){
        return num- other.num;
    }


    pair(int a, int b)
    {
        num=a;
        idx=b;
    }
}