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 in = new Scanner(System.in);
		int n = in.nextInt();
		int[] array = new int[n];
		int[] b = new int[n];
		for(int i=0;i<n;i++) {
			array[i] = in.nextInt();
			b[i] = array[i];
		}
		in.close();
		Arrays.sort(b);
		long m = (long)n*(n+1)/2;
		int l = 0;
		int r = n;
		int x = b[n/2];
		while(r-l>1) {
			x = b[(r+l)/2];
			int[] a = new int[n];
			for(int i=0;i<n;i++) {
				if(array[i]>=x) a[i] = 1;
				else a[i] = -1;
			}
			int[] sum = new int[n+1];
			int max = 0;
			int min = 0;
			for(int i=0;i<n;i++) {
				sum[i+1] = sum[i] + a[i];
				if(sum[i+1]>max) max = sum[i+1];
				if(sum[i+1]<min) min = sum[i+1];
			}
			long count = 0;
			int[] bit = new int[max-min+1];
			for(int i=0;i<n+1;i++) {
				count = count + bitGetSum(bit, sum[i]-min);
				bitAdd(bit, sum[i]-min, 1);
			}
			if(count>=(m+1)/2) l += (r-l)/2;
			else r -= (r-l)/2;
		}
		System.out.println(b[l]);
	}
	
	
	public static long bitGetSum(int[] bit,int idx) {
		idx++;
		long ans = 0;
		for(;idx>0;idx-=(idx&-idx)) {
			ans = ans + bit[idx-1];
		}
		return ans;
	}
	
	public static void bitAdd(int[] bit,int idx,int num) {
		idx++;
		for(;idx<=bit.length;idx+=(idx&-idx)) {
			bit[idx-1] += num;
		}
	}

}
